180asolns7 - 180A HW 6 Solutions 1 Suppose X has the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 180A HW 6 Solutions November 30, 2006 1) Suppose X has the exponential distribution with parameter and Y has the exponential distribution with parameter . Assume X and Y are independent. Let Z = X/(Y + 1). (a) Compute E[Z]. (b) Compute the PDF of Z. Answer (a) Since X and Y are independent, X and 1/(Y+1) are independent, and we have E[Z] = E[X/(Y + 1)] = E[X]E[1/(Y + 1)] = (1/)( 0 1 e-y dy). y+1 Note that E[1/(Y + 1)] = 0 (1/y + 1)e-y dy does not have a closed form solution. (b) We will find the PDF of Z by finding the CDF of Z, G(z), and then differentiating it: z(y+1) G(z) = P (Z z) = P (X/(Y + 1) z) = P (X z(y + 1)) = 0 0 f (x, y)dxdy Where f(x,y) is the joint density of X and Y . Since X and Y are independent, f (x, y) = fX (x)fY (y) = e-x-y . Thus, z(y+1) G(z) = 0 0 0 (e-x - y)dxdy = (-e-z(y+1)-y + e-y )dy = -e-z(y+1)-y e-y - ] = -z 0 e-z 1- z [ Thus the PDF of Z, g(z), is: g(z) = G (z) = e-z - ze-z z 2 1 2) Let X and Y be random variables with PDF f (x, y) = x + y, x, y (0, 1), and f (x, y) = 0, otherwise Answer 1 1 1 1 0 E[X] = 0 0 xf (x, y)dxdy = 0 1 0 (x2 + xy)dxdy = (1/3 + y/2)dy = 7/12. E[X 2 ] = 0 1 0 1 x2 (x + y)dxdy = 0 1 0 1 0 1 (x3 + x2 y)dxdy = (1/4 + 1/3y)dy = 5/12. V ar[X] = E[X 2 ] - E[X]2 = 5/12 - (7/12)2 = 11/144. 1 1 1 1 0 1 0 E[XY ] = 0 0 xy(x + y)dxdy = 0 (yx2 + xy 2 )dxdy = (y/3 + y 2 /2)dy = 1/3. Cov(X, Y ) = E[XY ] - E[X]E[Y ] = 1/3 - (7/12)(7/12) = -1/144. Since our density is symmetric in X and Y , E[Y ] = E[X] = 7/12 and V ar[Y ] = V ar[X] = 11/144. Thus E[Z] = E[X] + E[Y ] = 7/12 + 7/12 = 7/6 and V ar[Z] = V ar[X + Y ] = V ar[X] + V ar[Y ] + 2Cov[X, Y ] = 11/144 + 11/144 - 2/144 = 20/144 = 5/36. 3) Suppose X1 , X2 are independent, each with the uniform distribution on (0, 1). Compute the probability density function of Y = X1 + X2 . Answer The convolution of X and Y gives: fZ (t) = - fX (x)fY (t - x)dx. Since 0 fX (x) 1 and 1 - t fY (t - x) t for X, Y uniform on [0, 1], we have: t fZ (t) = 0 1 dx for 0 t 1, dx for 1 t 2, t-1 fZ (t) = fZ (t) = 0 for t > 2 or t < 0. 2 Thus 0 t fZ (t) = 2-t 0 t0 0t1 1<t2 t>2 4) Show by induction (on n) that Y = X1 + ... + Xn has the negative binomial distribution with parameters (n, p) when the Xi s are independent, each with the geometric distribution with parameter p. Answer For the base case n=1, note that N egBin(1, p) Geom(p). Now assume Z = X1 +...+Xn -1 N egBin(n - 1, p), and let Xn Geom(p). We want to show Z + Xn N egBin(n, p). By the convolution formula, P (Y = k) = P (Z + Xn = k) = all x t-1 x=n-1 P (Z = x)P (Xn = k - x) = x - 1 n-1 p (1 - p)x-(n-1) p(1 - p)k-x-1 = n-2 k-1 pn (1 - p)k-n x=n-1 x-1 n-2 From the hint, letting y = x - 1 we get k-1 x=n-1 x-1 n-2 k-2 = y=n-2 y n-2 = k-1 n-1 So P (Y = k) = Thus Y N egBin(n, p). 5) Let X1 , ..., Xn be independent, with Xi having the exponential distribution with parameter i . Define Y = n Xi . Compute E(Y k ) for k = 1, ..., 4. i=1 Answer First we will calculate E[Xin ], n = 1, 2, 3, 4: E[X n ] = 0 k-1 n p (1 - p)k-n n-1 xn e-x dx Using intregation by parts, we get: 3 E[X n ] = ([ -xn e-x ]0 ) + n 0 nxn-1 -x e dx = xn-1 e-x dx = n E[xn-1 ] 0 Since E[X] = 1/, 2 E[X] = 3 E[X 3 ] = E[X 2 ] = 4 4 E[X ] = E[X 3 ] = E[X 2 ] = Also, observe that Y 2 is: n 2 2 6 3 24 4 Y = (X1 + ... + Xn ) = i=1 2 2 Xi2 + i=j Xi Xj Y 3 is: n Y = (X1 + ... + Xn ) = ( i=1 n 3 3 Xi2 + i=j Xi Xj )(X1 + ... + Xn ) = Xi Xj )(X1 + ... + Xn )) = (( i=1 n Xi2 )(X1 + ... + Xn )) + (( i=j 2 Xi Xj ) + ( i=j i=j n ( i=1 Xi3 + Xi2 Xj + i=j 2 Xi Xj + i=j=k 2 Xi Xj + i=j Xi Xj Xk ) = Xi Xj Xk i=j=k Xi3 + 3 i=1 and Y 4 is: 4 n Y 4 = (X1 + ... + Xn )4 = ( i=1 n Xi3 + 3 i=j 2 Xi Xj + i=j=k Xi Xj Xk )(X1 + ... + Xn ) = Xi Xj Xk )(X1 + ... + Xn ) = 3 Xi Xj ) + ( i=1 n Xi3 )(X1 + ... + Xn ) + (3 i=j 3 Xi Xj ) + (3 i=j i=j 2 Xi Xj )(X1 + ... + Xn ) + ( i=j=k 2 Xi Xj Xk + 3 i=j=k i=j ( i Xi4 + 2 Xi2 Xj + 3 Xi Xj Xk Xl + i=j=k=l Xi2 Xj Xk i=j=k n + i=j=k 2 Xi Xj Xk + i=j=k 2 Xi Xj Xk = Xi4 + 4 i i=j=k 3 Xi Xj + 3 i=j 2 Xi2 Xj + 6 i=j=k 2 Xi Xj Xk + i=j=k=l Xi Xj Xk Xl Using these facts, along with the fact that expectation is a linear function and the Xi s are independent, we get E[Y ] is: n n n E[Y ] = E[ 1 Xi ] = 1 E[Xi ] = 1 1/i E[Y 2 ] is: n n E[Y ] = E[ i=1 2 Xi2 + i=j Xi Xj ] = i=1 2/2 + i i=j 1/(i j ) E[Y 3 ] is: n E[Y ] = E[ i=1 n 3 Xi3 + 3 i=j 2 Xi Xj + i=j=k Xi Xj Xk ] = 1/(i j k ) i=j=k 3/3 + 3 i i=1 i=j 2/(i /2 ) + j and E[Y 4 ] is: n E[Y 4 ] = i 4/4 + 4 i i=j=k 3/(i 3 ) + 3 j i=j 4/(2 2 ) + 6 i j i=j=k 2/(i j 2 ) + k i=j=k=l 1/(i j k l ) 5 ...
View Full Document

This homework help was uploaded on 02/03/2008 for the course MATH MATH 180A taught by Professor Castro during the Fall '08 term at UCSD.

Ask a homework question - tutors are online