Stochastic Calculus - Problem set 1 - Fall 2002

Exercise 1 - a We have 600 blue balls and 300 red balls, therefore the probability to choose one blue ball is \frac{600}{300+600} = \frac{2}{3}. Since every time we return the chosen ball, the probability that the first n balls are blue is:

(\frac{2}{3})^{n}

Exercise 1 - b First we notice that N, the number of blue balls chosen before the first red one, takes its values in the non-negative integers {0, 1, 2, ...}. If N = n, we choose n blue balls before the first red one, each of them with the probability \frac{2}{3}, therefore:

P (N = n) = (\frac{2}{3})^{n} \frac{1}{3}

Since for |x| < 1 the series ∑_{n=0}^{∞} x^{n} = \frac{1}{1−x} is uniformly convergent, so are all its derivatives, and we can take two successive derivatives to get:

∑_{n=0}^{∞} nx^{n} = \frac{x}{(1 − x)^{2}}

and

∑_{n=0}^{∞} n^{2}x^{n} = x\frac{1 + x}{(1 − x)^{3}}

If we plug in x = \frac{2}{3} we obtain E(N) = 2 and E(N^{2}) = 10. Since V ar(N) = E(N^{2}) − E(N)^{2}, the variance of N is V ar(N) = 6.

Exercise 1 - c Since N only takes discrete integer values, P (N ≤ 2) = P (N = 0) + P (N = 1) + P (N = 2) = \frac{1}{3} (1 + \frac{2}{3} + (\frac{2}{3})^{2}). But P (N = 0|N ≤ 2) = \frac{P (N=0)}{P (N≤2)}, and it follows P (N = 0|N ≤ 2) = \frac{9}{19}.

Exercise 1 - d The probability that N is an even number is:

∑_{k≥0} P (N = 2k) = \frac{1}{3} ∑_{k≥0} (\frac{2}{3})^{2k} = \frac{3}{5}

Exercise 2 - b Since ”Good” and ”Bad” are a partition of the universe Ω, we have P (L) = P (L|G)P (G) + P (L|B)P (B). We already know that P (L|G) = 0.7 and P (G) = 0.1, but P (L|B) = 1 − P (D|B) = 0.2 and P (B) = 1 − P (G) = 0.9. From this we deduce P (L) = 0.7 ∗ 0.1 + 0.2 ∗ 0.9 = 0.25.

Exercise 2 - c We can write P (G ∩ L) in two ways: P (G ∩ L) = P (G|L)P (L) = P (L|G)P (G) and thus P (G|L) = P (L|G)\frac{P (G)}{P (L)} = 0.7\frac{0.1}{0.25} = 0.28.

Exercise 3 - a The mean of this random variable is given by:

E(X) = 2 ∫_{0}^{1} x(1 − x)dx = \frac{1}{3}

and from the second moment,

E(X^{2}) = 2 ∫_{0}^{1} x^{2}(1 − x)dx = \frac{1}{6}

we deduce the variance V ar(X) = E(X^{2}) − E(X)^{2} = \frac{1}{18}.

Exercise 3 - b I see at least 2 ways to do this:

1) The characteristic function of the sum is the product of the characteristic functions, by independence, and what corresponds to a product in the Fourier space is a convolution, therefore:

f_{Y} (y) = ∫_{ℝ} f_{X_{1}}(x)f_{X_{2}}(y − x)dx

2) It is possible to make a linear change of variables such as Y = X_{1} + X_{2} and Z = X_{2} to find the joint density of the couple (Y, Z), then integrate over Z to get the marginal PDF for Y. Whichever method you choose, we end up with having to calculate the following integral:

f_{Y} (y) = 4 ∫ (1 − x)(1 − (y − x))111_{0≤x≤1}111_{0≤y−x≤1}dx

Simple algebra gives 111_{0≤y−x≤1} = 111_{y−1≤x≤y}, and therefore we have to look at 2 separate cases.

1) If 0 ≤ y ≤ 1 then 111_{y−1≤x≤y}111_{0≤x≤1} = 111_{0≤x≤y} and the integral is:

f_{Y} (y) = 4 ∫_{0}^{y}(1 − x)(1 − (y − x))dx = 4 (y(1 − y) + \frac{y^{3}}{6} )

2) If 1 ≤ y ≤ 2 then 111_{y−1≤x≤y}111_{0≤x≤1} = 111_{y−1≤x≤1} and the integral is:

f_{Y} (y) = 4 ∫_{y−1}^{1}(1 − x)(1 − (y − x))dx = −\frac{2}{3}(y − 2)^{3}

If we are in either situation y ≤ 0 or y ≥ 2, the integral is zero.

Exercise 3 - c By linearity of the expectation, we have E(Y) = E(X_{1} + X_{2}) = E(X_{1}) + E(X_{2}) = 2E(X) = \frac{2}{3}. Since X_{1} and X_{2} are independent, the variance of their sum is the sum of their variance, V ar(Y) = 2V ar(X) = \frac{1}{9}.

Exercise 3 - d We use the expression we found in question b) to calculate

P (Y > 1) = −\frac{2}{3} ∫_{1}^{2}(y − 2)^{3}dy = \frac{1}{6}

Exercise 3 - e If we denote by S_{100} = X_{1} + X_{2} + ... + X_{100} we want to estimate P (S_{100} > 34) using the Central Limit Theorem. We substract the expectation and normalize, to find the quantity we want to calculate is the same as P (\frac{S_{100}−100E(X)}{σ\sqrt{100}} > \frac{34−100E(X)}{σ\sqrt{100}}), where σ is the standard deviation of X.

The Central Limit Theorem says that the left handside converges in distribution to a Gaussian with mean 0 and variance 1, and according to the results of question a), we have E(X) = \frac{1}{3} and

σ = \frac{1}{3\sqrt{2}}. If Φ(x) = \frac{1}{2π} ∫_{−∞}^{x} e^{− \frac{z^{2}}{2}} dz is the CDF of a standard Gaussian, then the probability we want to estimate is 1 − Φ(0.2828) = 0.389.

Exercise 4 - a We have:

P (X^{2} + Y^{2} ≤ 1) = \frac{1}{8π} ∫∫_{x^{2}+y^{2}≤1}(4 − (x^{2} + y^{2}))dxdy

The fact we integrate over a circular domain, and the function explicitely depends on (x^{2} + y^{2}) indicates it may be a good idea to switch to polar coordinates. If we write x = r cos θ and

y = r sin θ, then the Jacobian matrix of the transformation is (\begin{matrix}cos θ & −r sin θ \\ sin θ & r cos θ \end{matrix}) whose determinant

is r, and therefore formally we have dxdy = rdrdθ. Thus we can write the above integral as:

P (X^{2} + Y^{2} ≤ 1) = \frac{1}{8π} ∫_{0}^{1} ∫_{0}^{2π} r(4 − r^{2})dθdr = \frac{7}{16}

Exercise 4 - b For a given x, it follows from the constraint x^{2} + y^{2} ≤ 4 that |y| ≤ \sqrt{4 − x^{2}}. Therefore if for a given x ∈ [−2; 2] we integrate with respect to y, we get:

f_{X}(x) = \frac{1}{8π} ∫_{−\sqrt{4−x^{2}}}^{√\overline{4−x^{2}}}(4 − (x^{2} + y^{2}))dy = \frac{1}{6π} (4 − x^{2})^{\frac{3}{2}}

Exercise 4 - c By definition we have Cov(X, Y) = E(XY) − E(X)E(Y). The joint density, both as a function of x or as a function of y alone is even. Therefore its expectation over a symmetrical domain is 0, and we have E(X) = E(Y) = 0. The same kind of argument proves that this is still true for E(XY), but to be convinced we can do the simple calculation:

E(XY) = \frac{1}{8π} ∫∫_{x^{2}+y^{2}≤4} xy(4 − (x^{2} + y^{2}))dxdy = \frac{1}{8π} ∫_{0}^{2} r^{3}(4 − r^{2}) ∫_{0}^{2π} cos θ sin θdθdr

and the integral with respect to θ is zero.

Exercise 4 - d If we look at the event P (X ≥ \sqrt{2}), it is easy to see that it is zero if Y ≥ \sqrt{2}, and it is not zero if Y < \sqrt{2}. Therefore X and Y are not independent. We know that the covariance of two independent random variables is zero, but except in the situation where the variables are jointly Gaussian, a zero covariance does not imply independence. This exercise provides us with a concrete example.

Exercise 4 - e If U = X^{2} and V = Y^{2}, then X = \sqrt{U} and Y = \sqrt{V}, and the determinant of the Jacobian matrix

(\begin{matrix}\frac{1}{2\sqrt{u}} & 0 \\ 0 & \frac{1}{2\sqrt{v}} \end{matrix} ) is \frac{1}{4\sqrt{uv}}. Now we have to find the image of the domain D = \{(x, y)|x^{2} + y^{2} ≤ 4\}

under the map u = x^{2},v = y^{2}. It is obvious that u ≥ 0, v ≥ 0, and u + v ≤ 4, and that every point of this triangle T has a preimage in the original disk. As a matter of the fact, except the origin, each point has 4 preimages: we have a onto map that is not into. In order to make our change of variables, we need to find a C^{1}-diffeomorphism. But if we restrict our original domain to any quarter of the disk, we have a one-to-one map and we can apply our theorem. Doing so on each quarter gives the same result, therefore in the resulting density, a factor 4 comes out of the disk decomposition. If we take any bounded continuous function f, we then have

E(f(U, V)) = \frac{1}{8π} ∫∫_{D} f(x^{2}, y^{2})(4 − (x^{2} + y^{2}))dxdy = 4\frac{1}{8π} ∫∫_{T} f(u, v)\frac{4 − (u + v)}{4\sqrt{uv}}dudv

But if this is true for any bounded continuous function f, then we must have:

f_{(U,V)}(u, v) = \frac{1}{8π} \frac{4 − (u + v)}{\sqrt{uv}}111_{(u,v)∈T}

Exercise 4 - f This is just a calculation, and I am not going to write every step of it. First we need to calculate

E(X^{2}) = \frac{1}{8π} ∫_{0}^{4} ∫_{0}^{4−u} u\frac{4 − (u + v)}{\sqrt{uv}} = \frac{2}{3}

and by symmetry we have E(Y^{2}) = \frac{2}{3} as well. Now we can switch to polar coordinates to calculate the integral:

E(X^{2}Y^{2}) = \frac{1}{8π} ∫∫_{x^{2}+y^{2}≤4} x^{2}y^{2}(4 − (x^{2} + y^{2}))dxdy = \frac{1}{3}

It follows that Cov(X^{2}, Y^{2}) = E(X^{2}Y^{2}) − E(X^{2})E(Y^{2}) = −\frac{1}{9}. If X and Y were independent, X^{2} and Y^{2} would be independent as well, and we would have Cov(X^{2}, Y^{2}) = 0. The fact it is not zero gives another proof of their non independence.