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

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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 ...
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