Sample Exam 2 Solutions - Math 464 - Fall 14 -Kennedy 1. Let X and Y be independent random variables. They both have a gamma distribution with mean 3 and variance 3. (a) Find the joint probability density function (pdf) of X, Y . Solution: Since they are independent it is just the product of a gamma density for X and a gamma density for Y . For the gamma distribution, µ = w/λ, σ 2 = w/λ2 . Since the mean and variance are both 3, λ = 1 and w = 3. So 1 x2 y 2 e−x−y if x ≥ 0, y ≥ 0 fX,Y (x, y) = Γ(3)2 0, otherwise (b) Express P (3X + Y ≤ 3) as an integral. Do not try to do the integral. Solution: The region where 3x + y ≤ 3, x ≥ 0, y ≥ 0 is the triangle in the upper right quadrant below the line y ≤ 3 − 3x. So we get Z 1 Z 3−3x 1 2 2 −x−y xy e dy dx Γ(3)2 0 0 2. Let X have an exponential distribution with E[X] = 1. Let Y = X 2 − 2. (a) Find the mean and variance of Y . Solution: First we compute some moments of X for later use. The mgf for X is m(t) = 1/(1 − t). 1 , E[X] = m′ (0) = 1, (1 − t)2 2 , E[X 2 ] = m(2) (0) = 2, m(2) (t) = 3 (1 − t) 6 , E[X 3 ] = m(3) (0) = 6, m(3) (t) = (1 − t)4 24 m(4) (t) = , E[X 4 ] = m(4) (0) = 24 (1 − t)5 m′ (t) = Now E[Y ] = E[X 2 ] − 2 = 2 − 2 = 0 E[Y 2 ] = E[(X 2 − 2)2 ] = E[X 4 − 4X 2 + 4] = 24 − 4 · 2 + 4 = 20 1 So var(Y ) = 20 − 0 = 20. (b) Find the probability density function (pdf) for Y . Solution: We start by finding the cdf for Y . FY (y) = P (Y ≤ y) = P (X 2 − 2 ≤ y) = P (X 2 ≤ y + 2) Z √y+2 p p e−x dx = 1 − exp(− y + 2) = P (X ≤ y + 2) = 0 Take the derivative of this to get p 1 fY (y) = (y + 2)−1/2 exp(− y + 2), y ≥ −2 2 The range for Y is [−2, ∞). 3. Let X and Y be continuous random variables with joint pdf 3 fX,Y (x, y) = (x2 + y 2 ), 2 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 Outside of 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, fX,Y (x, y) = 0. (a) Find the marginal densities of X and Y The marginal density of X for 0 ≤ x ≤ 1 is Z 1 Z 1 3 3 2 1 1 3 3 2 2 y 2 dy] = [x2 + ] = + x2 (x + y ) dy = [x + fX (x) = 2 2 3 2 2 0 0 2 So fX (x) = 1 2 + 32 x2 if 0 ≤ x ≤ 1 otherwise 1 2 + 23 y 2 if 0 ≤ y ≤ 1 otherwise 0 The same calculation shows fY (y) = 0 (b) Are X and Y independent? Solution: They are not independent since fX,Y (x, y) is not equal to fX (x)fY (y). 2 4. Let X, Y be jointly continuous random variables with joint probability density function (pdf) 4xy, if 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 fX,Y (x, y) = 0, otherwise Let Z = X + Y . Compute fZ (z), the probability density function (pdf) for Z. Solution: The range of X, Y is the unit square. The range of Z will be [0, 2] We need to compute the cdf, P (Z ≤ z) = P (X + Y ≤ z). How the line x + y = z intersects the unit square depends on whether 0 ≤ z ≤ 1 or 1 ≤ z ≤ 2. In the first case Z z Z z−x 4xy dy dx P (X + Y ≤ z) = 0 0 After some calculation this equals 61 z 4 . For 1 ≤ z ≤ 2, P (X + Y ≤ z) = Z z−1 0 Z 1 4xy dy dx + 0 Z 1 z−1 Z z−x 4xy dy dx 0 After an unreasonable amount of calculation this equals − 61 z 4 + 2z 2 − 83 z + 1. So the pdf is 23 z 3 , if 0 ≤ z ≤ 1 2 8 3 fZ (z) = − 3 z + 4z − 3 , if 1 ≤ z ≤ 2 0, otherwise 5. Let X and Y be independent random variables, each of which has a standard normal pdf. Let Z = Y − X + 4. (a) Find the mean and variance of Z. Solution: E[Z] = E[Y ] − E[X] + 4 = 4. var(Z) = var(Y ) + var(−X) = var(Y ) + var(X) = 1 + 1 = 2. (b) Find the probability density function (pdf) of Z. Hint: this can be done with very little computation. Solution: It is easy to show that −X is also a standard normal. The sum of independent normal random variables is normal, and adding a constant to a normal random variable gives another normal random variable. So Z is 3 normal. Part (a) tells us its mean and variance. Another way to see this is to look at the mgf. Since X and Y are independent, functions of them are independent. So MZ (t) = E[exp(t(Y − X + 4))] = e4t E[exp(tY )]E[exp(−tX)] 1 1 = exp(4t + t2 + (−t)2 ) = exp(4t + t2 ) 2 2 which is mgf of a normal with mean 4 and variance 2. So 1 1 exp( (x − 4)2 ), fZ (z) = √ 4 4π −∞ < z < ∞ 6. Random variables X and Y have joint cumulative distribution function (cdf) 1 [ π tan−1 (x) + c](1 − e−y ), if y ≥ 0 FX,Y (x, y) = 0, if y < 0 where c is some constant. (a) Are X and Y independent? Solution: Yes, the joint cdf factors into a function of x times a function of y, so they are independent. (b) Find the value of c. Solution: lim F (x, y) = x,y→∞ 1 1π +c= +c π2 2 This must equal 1, so c = 1/2. (c) Find the joint probability density function (pdf) for X, Y . Solution: We take the second order partial derivative of FX,Y (x, y) with respect to x and y. This gives 1 1 e−y , if y ≥ 0 fX,Y (x, y) = π 1+x2 0, if y < 0 Note that X, Y are independent. X has the Cauchy distribution, and Y is exponential with λ = 1. 4 7. The joint pdf of X and Y is −x−y 2e , if x ≥ 0, y ≥ 0, y ≥ x fX,Y (x, y) = 0, otherwise Define new random variables by U = Y −X √ X V = (a) Are X and Y independent? Solution: No, the condition y ≥ x does not factor. Another way to see they are not independent is to look at P (Y ≤ 1, X ≥ 2). If we compute this we will integrate the joint density over a region where it is zero, so P (Y ≤ 1, X ≥ 2) = 0.. But P (Y ≤ 1) and P (X ≥ 2) are both not zero. (b) Find the joint density of U, V . Solution: Solving for the inverse we get X = V2 Y = U +V2 So the Jacobian is J = det 0 2v 1 2v = −2v As function of u, v, f (x, y) becomes 2 exp(−u − 2v 2 ). So the joint density of U, V is 4v exp(−u − 2v 2 ). We need to determine the range of U, V . Clearly v ≥ 0. The condition y ≥ x implies u ≥ 0. So the range is contained in the region u ≥ 0, v ≥ 0. To see if all of the upper right quadrant is in the range we look at the inverse equations and ask if for any u ≥ 0, v ≥ 0 we get x, y satisfying x ≥ 0, y ≥ 0, y ≥ x. We do, so the range is all of u ≥ 0, v ≥ 0. So 4v exp(−u − 2v 2 ), if u ≥ 0, v ≥ 0 fU,V (u, v) = 0, otherwise (c) Are U and V independent? Solution: Yes, the joint pdf factors into a function of u times a function of v. 5 8. Let X and Y be independent random variables. X has an exponential distribution with E[X] = 2. Y has an exponential distribution with E[Y ] = 1. Let Z = X + 2Y . (a) Find the mean and variance of Z. ‘Solution: For X, the parameter is λX = 1/2 and for Y it is λY = 1. Using the formula sheet, Mean is E[Z] = E[X] + 2E[Y ] = 4. Variance is V ar(Z) = V ar(X) + 4V ar(Y ) = 8. (b) Find the moment generating function (mgf) of Z. Solution: X and 2Y are independent, so MZ (t) = MX (t)M2Y (t) = MX (t)MY (2t) = 1 1 1/2 = 1/2 − t 1 − 2t (1 − 2t)2 (c) The pdf of Z is in our catalog. What is it? Solution: The mgf is that of a gamma distribution with λ = 1/2 and w = 2. 6