Transcription
Hello, and welcome to another lecture of this course, Mathematics for Economics Part I. Now, the module that we have been covering is integration. So, we have talked about how integration can be interpreted as area under a curve, and we talked about indefinite integrals; we also talked about definite integrals, and right now we are talking about certain exercises where integration and its tools can be used. And here, on your screen, you can see one particular exercise. So, this is an exercise of a definite integral. The question is: evaluate integration 1 to 2, and the integrand, that is the function we want to integrate, is 3x divided by 10.
Now, how to do that? As we have seen earlier, that if there is a constant in the integrand, that can be taken out. So, here the constant is 3/10; we take that out. So, inside you just have x. And integration of x dx, as we know, it's x squared divided by 2, and limits are 2 and 1. So, essentially what we are doing is 3 divided by 10 multiplied by half of 2 squared minus 1 squared, and that essentially boils down to 9 divided by 20. This was a simple problem of a definite integral.
Here is a problem from finance and economics. A theory of investment has used a function, capital W, defined for all capital T by WT, that is W(T), is equal to K divided by capital T and integration 0 to capital T e to the power minus alpha t dt. Here, capital K is a given constant, positive; alpha is also a given constant, positive. What we need to do: evaluate the integral and prove that W(T) takes values in the interval 0 to K and is strictly decreasing. So, there are three parts of this problem.
Now, let us try to understand what this is about. So, WT, in some sense, it measures the value of investment, and how this investment is measured, it is—remember, it is a function of capital T. Capital T is the total time over which the investment will give some return. So, that is the total time span. That is the interpretation of capital T. It is so WT is a function of capital T, but there are some complications here. There is an integral here, a definite integral. And you have e to the power minus alpha t, which has to be integrated over the interval 0 to capital T. We shall later on see that this particular thing, e to the power minus alpha t, integration of that has an interpretation of present discounted value, but that will come later on. The first thing that we have to do is that we have to evaluate this: W(T) is K divided by capital T integration of 0 to T e to the power minus alpha small t d small t.
Now, this is not very hard to integrate. Basically, you have an exponential function whose power is minus alpha t. Alpha is a constant. So, we know how to do that. It is e to the minus alpha t divided by the coefficient of small t, and the coefficient of small t here is minus alpha. So, that is coming in the denominator. And in this context, capital K and capital T they are sitting outside, because they are not functions of small t. And we have to take the limit value, at the limits, and minus e to the power minus alpha capital T minus of e to the power minus alpha multiplied by 0. And you have to be careful that there is a negative sign here. So, this basically boils down to this: capital K divided by capital T multiplied by alpha multiplied by 1 minus e to the power minus alpha capital T. As you can see, this expression is a function of capital T. That is why you have W as a function of capital T on the left-hand side. So, the first part is done.
Now, what we need to show is that W(T) takes the values in the interval 0 to capital K. What is the range of capital T? Depending on the range of capital T, W will take different values. Now, we have been given this information that capital T is not negative. It always takes a positive value. So, it basically varies between 0 and plus infinity. So, we can take these two limits and try to see how the value of the function changes—functions mean W(T)—how W(T) changes as capital T takes these two extreme values. That is what we have done here. As capital T goes to infinity, that is positive infinity, this is the form of the function. You can straightaway see that capital K divided by capital T this term goes to 0. So, the whole expression will go to 0. So, capital W(T) goes to 0 as capital T goes to infinity. Now, that is one extreme. What about the other extreme? The other extreme is: suppose capital T approaches 0, the minimum value; we know that capital T is positive. So, it can go as close to 0 as possible from the right-hand side. Now, so we have to evaluate this, to be more precise, T tends to 0 plus. Now, this we can evaluate by applying L'Hopital's rule. So, we differentiate both the numerator and the denominator in this expression with respect to capital T. And if we do so, basically, W(T) becomes this expression, which is capital K multiplied by e to the power minus alpha T. Now, we can take the limit. And if we do that, as T goes to 0, then this term becomes 1. It goes to 1 because e to the power of 0 is 1. So, the value of the function approaches capital K as T goes to 0. And that was our second task. Because what was the second task saying? That this function takes values from this range, 0 to K. That is what we have shown here. If T goes to 0, the value of the function approaches K. And as T goes to infinity, the value of the function approaches 0.
Now, we are also given the information that K is positive, and as we can see that if K is positive, then as T increases, it seems that the function is declining in value. It started from K, which is positive, and it goes to 0 as T goes to infinity. Now, we are not sure whether it is a monotonically decreasing function or not. For that, to show that—and that is the third task actually, that it is strictly decreasing—so, for that I have to take the derivative of the function, that is, the derivative of the W(T) function with respect to capital T. So, that is what we have done here, and we simplify this expression and simplify to this: K multiplied by e to the power minus alpha capital T divided by alpha capital T squared, whole thing multiplied by minus of e to the power alpha capital T plus 1 plus alpha T. Now, this expression turns out to be negative. The reason being, e to the power alpha T is always greater than 1 plus alpha T. So, this is true: e to the alpha T is strictly greater than 1 plus alpha T, which basically means that this expression is negative, and that essentially proves that the function, that is W(T), is declining in T. So, that was our third task.
Now, we come to some applications of integration, in particular, definite integrals. We will be concentrating more on the applications of definite integrals. Suppose at time 0, t is equal to 0, extraction from an oil well—so this is a problem of extraction of oil from an oil well—it starts extracting at time 0. The well contains capital K barrels of oil at that point of time. That is, at t is equal to 0, the oil well contains capital K barrels of oil. Let us suppose x(t), so it is a function of small t, is the amount of oil in barrels that is left in the well at the time small t. So, think about this in terms of a diagram. Here you have t is equal to 0. Now, here how much oil is there? Suppose this is capital K. Now, and take another time, let us suppose generically small t is the time; now, here what is the amount of oil that is left? Let us denote that by x(t). So, this is the value of the oil or the amount of oil that is left in that oil well, as let us suppose this is denoted by small x(t). Small x(t) is a decreasing function of t. How much oil will be left in the oil well that goes on declining, because extraction is taking place at each point of time. To begin with, when time was 0, the total amount was K. So, I can write K is equal to x(0). So, you can think of this total amount of oil left in the well to be a declining function of time.
What is the extraction per unit of time? So, how much oil is being extracted per unit of time? That can be written as minus of x(t + Δt) minus x(t) divided by Δt. Why is that? The reason is that how much oil is extracted per unit of time. If I assume this extraction rate of extraction to be positive, then notice x(t) minus x(t + Δt) divided by Δt is always positive, because with time the value of the x(t) is declining, so this term will be positive. So, this is going to be a positive expression. And so the rate of extraction can be given as minus of… Now, if x as a function of t is differentiable, and as t goes to 0, the above expression approaches to minus of x'(t), because this is nothing but the Newton quotient. Now, let us denote u(t) as the rate of extraction at point t. Therefore, we can write this as u(t) is equal to minus of x'(t), where x(0) is equal to K. Note, u(t) is a flow variable, whereas x(t) is a stock variable. U(t) is the change in the stock of oil, and that stock of oil is given by x(t). U(t) is a flow, and x(t) is a stock. The above two relations can be summarized as follows: The x(t), or the amount of oil that is left in the well at point t, is equal to capital K minus integration 0 to small t u(θ) dθ.
What is the interpretation of this? The amount of oil left at time small t, that is x(t), is equal to the initial amount, which is capital K, minus the total amount that has been extracted in the period 0 to t. So, that is given by this: the total amount of oil that has been extracted from the well in the period 0 to t. Remember, u(θ) is the rate of extraction at point θ. So, if I take the integration of 0 to t, that will give me the total amount of extraction from the point 0 to the point small t. So, we can take a very simple example where suppose the rate of extraction is constant and it is given by, let us suppose, ū, then we can play around with the above expression, this expression. So, in this case, u(θ) is equal to constant; it is ū. So, I can replace that in this formula. And if I do so, ū will—it can be taken out; it is a constant. So, if I integrate 1, it becomes only θ. And so this becomes capital K minus ū small t. So, this is the total amount of oil left in the well after t amount of time. And this is quite intuitive, because at each time, at each point of time the rate of extraction is ū. So, after t point of time, the total extraction will be ū multiplied by small t.
Now, one question that can be asked is: at what point of time the oil well becomes empty? Now, I can solve that by taking x(t) equal to 0 and solve for this t. We want to find the time in which the stock will deplete to 0. And the stock we know is x(t). So, that is why we are setting x(t) is equal to 0. And I use this formula here. So, K minus ū small t is equal to 0, and that is equivalent to saying that t is equal to K divided by ū. This expression that we are getting here, it is giving me the time at which the oil well will run dry. Now, this could have been done directly also, because the total stock of oil is K, and per unit of time ū is being extracted, which is constant. So, the time that will be required to extract all the oil from the oil well will be K divided by ū.
Now, this problem of oil extraction can be extended in a different context. So, suppose capital F as a function of small t is the foreign exchange reserve of a country at time t. So, what is foreign exchange reserve? Every country has a stock of currencies which are issued by other countries. So, for example, India will be having a stock of, let us say, US dollars. US dollars are the most accepted currency all over the world. So, India, whatever the foreign exchange reserve it has, will be mostly denominated by US dollars. But there could be other currencies also, like euro or other currencies like the Chinese currency or the Japanese currency. So, all those countries' currencies will be held as stock by India, and that stock is called the foreign exchange reserve. Likewise, other countries also have their own foreign exchange reserves. Now, this F, which is a function of t, so it basically means that with time the foreign exchange reserve can go on changing. If it is a differentiable function, then the rate of change in the reserve per unit of time is, let us suppose, given by small f(t), which is equal to capital F'(t). Now, if small f(t) is greater than 0, it means there is a net inflow of foreign exchange in the country at the time t. And if f(t) is less than 0, it means there is net outflow. So, for example, if f(t) is less than 0, that means that foreign exchange reserve, which is capital F(t), is depleting. On the other hand, if a small f(t) is greater than 0, then the foreign exchange reserve is accumulating in value.
Now, notice, I said that this example is similar to the oil well example. It is similar, but not exactly the same. In the case of an oil well, the extraction goes only in one direction. So, as long as the rate of extraction is positive, the stock of oil that is there, which was x(t), will always be declining. But here, capital F(t), which is the stock variable here, can increase if small f(t) is positive. It can also decrease if there is a depletion of the foreign exchange, which occurs if small f(t) is negative. Now, over the period t1 and t2, the change in the foreign exchange reserve is given by capital F(t2) minus capital F(t1), and that we have seen is it can be expressed in the form of an integration. It is a definite integral. So, integration of t1 to t2 and the integrand is small f(t) dt. In this diagram, the change in the foreign exchange reserve is positive between t1 and t2. So, in this diagram, notice what we have done here: along the horizontal axis I have time, and along the vertical axis, mind you, I do not have capital f(t), I have small f(t). And the area under the curve—remember that is the idea—the area under the curve, this curve that is small f(t), if I take two limits, here suppose the limits are t1 and t2, then the area under this curve, f(t), will denote how much the foreign exchange reserve is changing over this particular time frame. Now, in this diagram, I have drawn the diagram in such a way that this area, which is above the horizontal axis, I have shaded that, is more than in value than these areas which are below the horizontal axis. So, the area above the axis, above the t axis, these are the increments in the exchange—that is, foreign exchange reserves. And these are what? These are depletion. Now, the picture has been drawn in such a manner that the total amount of increment is more than the total amount of depletion. This area, the slanted shades of that region, is more than the horizontal shades region. So, that is why I am saying that the foreign exchange reserves change over t1 and t2 is positive. However, if smaller periods like t1 and t' are taken, then the change in the reserve is negative. So, that is quite obvious. So, I take the period from t1 to t'; so this small period. Here you can see that there is this area which is below the t axis. And if it is below the t axis, that means small f is negative. And therefore, there is depletion in foreign exchange in this particular timeframe.
Now, we come to a third application of this idea of definite integrals. Let capital F(r) be the proportion of individuals in a country with income less than or equal to r rupees per year. So, r here is income per year, and capital F(r) is the proportion of individuals in a country with income less than or equal to r rupees per year. So, to give you an example, suppose someone says that in India 60 percent of India's population have an income which is less than 1 lakh rupees per year. Suppose that is the statement. So, in that case, here F(r) will be like 0.6, 60 percent, and r is 1 lakh. The second point is this: If small n is the total number of individuals, then n multiplied by F(r) is the number of people with income no greater than r. So, this is the second statement. Let us try to understand this statement in terms of that example I have just given. I said 60 percent of India's population have income less than 1 lakh rupees. Now, I did not say how many people in India earn less than 1 lakh rupees. To get that number, how many people earn income less than 1 lakh rupees, suppose that number I have to find out. Now, how do I find that out? I have to multiply 60 percent with the total population of India, and that is what I have done here. Let us suppose India's population is—let us take an arbitrary figure—let us suppose this is 100 crore people is India's population. So, 100 crore multiplied by 60 percent, that means 60 crore people in India earn income less than 1 lakh rupees. So, by this manner, we can actually find out how many people are earning less than a particular level of income.
Thirdly, let us suppose r0 and r1 be the lowest and the highest income levels. So, you can imagine there is a minimum income level below which people do not earn; so that is r0. And in the other extreme, there could be a maximum or the highest income among all the income levels in a country, and that highest income level let us denote that by r1. So, all the incomes in the country will lie between these two extreme values, r0 and r1. Now, in general, this capital F(r) is a discontinuous function, because rupees are practically discontinuous. However, it approximates a continuous function if there are a large number of individuals. So, this is quite practical that r is what? It is income level. And income levels are measured in terms of the currency of the country. In India's case, it is the Indian rupee. But income levels are not continuous. It could be, let us suppose, 1000 rupees or 1001 rupees, 1002 rupees. So, there are gaps between 1000 and 1001. So, it seems that this F(r) function is likely to be a discontinuous function. But one can imagine that there are a large number of people. In India's case, it is close to 140 crore people. That is a huge number. So, one can actually say that this function approximates to a continuous function. Let us assume a particular function which is small f(r), which is the derivative of capital F(r). And suppose this is valid for all levels of income in this range, r0 to r1. Then actually there are standard names for these two functions: small f(r) and capital F(r). They are called income density functions—that is, small f(r). And capital F(r) is called the associated cumulative distribution function. So, one is density; the other is distribution.
Now, let us pick up two income levels, a and b, from this range, that is r0 to r1, then the proportion of individuals in this range will be—this is very easy—it is definite integral a to b f(r) dr. f(r) is the income density function, so, and this was obtained by differentiating the distribution function. So, if I take the definite integral from a to b of the density function, then I get the proportion of individuals who earn between these two income levels, a to b. And how many people are there in that interval? So, like the previous case, I multiply this term with smaller n. Small n is the total number of people in the country. So, this is the total number of people earning income from a to b. But the next question is: how much income are these people earning who belong to this interval, a to b? So, that is an interesting question. What is the total income of these people? Let us assume the function which is M(r). M(r) is the total income of all people earning no more than r. r is a particular level of income. So, there will be many people who will be earning less than r. I want to measure their total income, that is being denoted by capital M(r). Let us now consider a change in the income from r to r + Δr. So, the change in the income is Δr. Now, there are approximately n multiplied by f(r) multiplied by Δr people in that interval. How am I saying that? f(r) multiplied by Δr is the proportion of people, approximately—this is the proportion of people in that Δr interval. And I am multiplying that by n. So, n multiplied by f(r) multiplied by Δr is the number of people who belong to that interval. And what is their combined income? So, these numbers of people are there, but what is their total income? I can multiply that with small r, because I am taking the value at r. So, I can multiply that with the level of income, which is r, and that will give me approximately the total income earned by these people in that interval, Δr interval. In other words, I can write this: M(r + Δr) minus M(r). What is the left-hand side? It is nothing but the same thing that we are talking about. It is the total income of the people in that interval, Δr interval. It is M(r + Δr) minus M(r). And that is what we have just discussed. It is roughly equal to r multiplied by n multiplied by f(r) multiplied by Δr. Now, we can divide both sides by Δr, then the left-hand side becomes the Newton quotient. And if I take the Newton quotient and take the Δr approaching 0, then I get M'(r) on the left-hand side. So, from here I can get M(r + Δr) minus M(r) divided by Δr, and limit Δr going to 0 on the right-hand side, it is simply rn and f(r), and this is nothing but M'(r). So, that is what I have written here: M'(r) is equal to nr multiplied by f(r). Or the total income of individuals earning between the limits b and a is this, because this is the derivative. So, remember, if I take the integration, definite integral of the f'(function), that is the differentiated function which is M'(r) here, then I get the total income of the individuals between that interval. So, that is what I have done here. I have taken the integral of a to b of this function; n will come out because n is a constant, and what is the variable of integration? It is r, because r is changing. That is the income level is changing. So, we have found out the total income of individuals earning income between two limits, a to b. It is given by n multiplied by integration of a to b r f(r) dr. Therefore, the mean income of individuals earning between these two income levels is given by this. So, essentially what I have done is that I have taken this, which is the total income of the individuals earning between a to b, and divided it by the total number of people in that interval. And that number of people, this denominator, is a to b f(r) dr, that we have just seen before. So, n and n will get cancelled from numerator and denominator, and we are going to get this expression. So, this is the mean income of individuals earning between two limits, a to b. And often used income density function is called the Pareto distribution, given by f(r). So, this is the Pareto. In this case, its density function is small f(r). It is given by capital B r to the power minus beta. Here…
Beta and B are positive constants. Usually, beta has a value between 2.4 and 2.6. So, this is an often applied density function in the study of income distribution, the Pareto distribution.
Here is a related application of the same idea: demand for a good and income distribution. Suppose the demand for a good in the economy is given by D(p, r). So, demand, that is D, is a function of p and r. It is p is price and r is income, like before, r is income, of the individuals of the economy. So, demand is only dependent on these two variables: the price of the commodity that we are talking about (that price is given by small p) and the income of the individuals. And let us suppose small fr is the income density function. Then it can be shown that the total demand generated by individuals with income between small a and small b is given by g, which is a function of small p, and it is equal to n multiplied by the integration from a to b of [D(p, r) * fr] dr.
So, look at the general pattern here: fr is occurring, Dr is occurring, but in front there is the demand function, which is a function of r. You can see the same pattern here also. Here we are talking about not demand, but total income of people who are earning between a and b. So, here also you have n * fr dr, but here r is occurring just before fr. In this case, it is the demand because we are not talking about the total income but we are talking about demand. So, fr is preceded by the demand function, which is D(p, r).
Now, we come to another, the fourth probably, application of integration: the present discounted value of a continuous income stream. Suppose that income is received continuously from t = 0 to t = T at the rate of ft rupees per unit of time at time T. So, per unit of time you are getting ft, and the total time in which you are going to receive is from 0 to T. The interest is compounded continuously at the rate of r. Let Pt be the present discounted value of all payments made over the period 0 to t. Now, Pt is the money one has to deposit at point 0 to match what results from continuously depositing the income stream ft over the time interval 0 to t. This is what we have talked about—what is the idea of present value. So, here the present value is denoted by T. So, it will give me some return over a period of time, and that return must match with the income stream of ft that I am talking about.
Now, the present value of the income received in t to t + dt (so I have taken a change of time from t to t + dt) is given by this: P(t + dt) - P(t). This is the change in the present value if I take two different points of time. That income to be obtained in the future, remember, this is a present value of some income to be obtained in the future. And what is that income that is going to be obtained in the future? It is approximately equal to ft; ft is the stream at point t, and you are changing the time a little bit, so this amount of income that you are going to earn in the future in that dt small period is approximated by ft * dt. And the present discounted value of ft * dt is what? It is ft * dt * e^(-rt). Thus, [P(t + dt) - P(t)] / dt (I have taken dt to the left-hand side) is approximately equal to ft * e^(-rt). And I can take dt going to 0, and so the left-hand side becomes P’(t); P’(t) = ft * e^(-rt). And so I can now talk about the total amount of money, the present value of that, that will be obtained in the future; it is given by P - P(0), and this will be nothing but the definite integral from 0 to T of [ft * e^(-rt)] dt.
The present discounted value evaluated at t = 0 of a continuous income stream at the rate of ft per year over the period 0 to T with continuously compounded interest rate r is given by this expression. So, this is the expression for present discounted value. Remember, we saw something similar in that exercise of investment that we discussed just now. If the same expression is valued at t = T rather than at 0, then we get FDV or the future discounted value. So, here Pt is the discounted value when we are standing at point 0 and we are evaluating the future income stream. But suppose that evaluation is done in the future at T, then what happens? Then it is called the future discounted value. And how is it evaluated? I just have to multiply the PDV by e^(r * T). So, in this case, FDV = e^(rt) * PDV, and that simplifies to the integration from 0 to T of [ft * e^(rT - t)] dt.
In a similar manner, the discounted value at any time t = s of a continuous income stream ft over s to T is given by DV (discounted value). So, here instead of standing either at 0 or at T, suppose a person is standing at s, and s is between 0 and T. Then how does she evaluate the future income stream? So that is given by this: It is the integration from s to T of [ft * e^(-r(t - s))] dt. You can verify that here if I change s to be equal to 0, I will get the PDV.
Here is an example: Find the PDV and FDV of a constant income stream of 1000 rupees per year over the next 10 years with an interest rate of 10 percent per annum compounded continuously. So, PDV: I just directly apply the formula: the integration from 0 to 10 (10 years are there) multiplied by the stream of income; in this case, it is constant, so ft is just 1000, e^(-rt); r, that is the rate of interest, is 10 percent, so it is -0.1 * t dt. And so the things are now falling into place. So, I just have to integrate and take the definite integral. It comes out to be roughly 6300 rupees. This is the PDV. For FDV, it is just converting the PDV into FDV. I have to multiply that with e^(r * T). So, e^(0.1 * 10) = e^1. So, e * PDV, and e is roughly 2.72, and that boils down to 17136. So, you can see that there is a huge change in the value depending on whether we are talking about PDV or FDV.
Let us do this one, and then we shall call it a day. Find the present discounted value of a constant income stream of a rupees per year over the next T years, assuming an interest rate of r annually compounded continuously. What is the limit of the PDV as T goes to infinity? So, there are two parts to this. First, we have to find the PDV. Let us do that. It is simple. I just have to plug in this information into the formula. Here ft = a, r^(-rt), and the time span is 0 to T. Now, this obviously is a; I have to evaluate that, but notice that this is a function of T. So, I can write this as PT, denoting the present discounted value, which is a function of T. And I do this integration, and I find this PT to be of this form: a / r * (1 - e^(-rT)). Now, the second part was this: as T goes to infinity, then what is the limit of this PDV? Well, I have to take the limit of PT as T goes to infinity, so this is what we are getting into, and so this will become just 0, because e^(-rT) as T goes to infinity will become 0. So, this becomes a / r. This expression is the same as the PDV of an infinite future income stream of a rupees per year discounted at r rate of interest per year, but not continuously. So, this we have talked about before when we were talking about geometric series. There I did not talk about continuous discounting. So, there was discounting, but the discounting was discrete. So, there also I got the same expression: a / r is the present value. So, here also we are seeing the same thing. The exact example that I gave was that our bond which pays return in perpetuity; there the value of the bond or the price of the bond was given by a / r.
I think I will stop here, and I think there will be just one more lecture that will be required to finish this topic of integration. So, I will do it in the next lecture. Thank you for joining us. Thank you.