Practice Midterm 2 Partial Solutions

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1 8.440 Practice Midterm Partial Solutions. (0 points) Let and Y be independent Poisson random variables with parameter. Compute the following. (Give a correct formula involving sums does not need to be in closed form.) (a) The probability mass function for given that + Y = 5. ANSWER: Write p(k) = P { = k} = e /k!. Then suppose x {0,,, 3, 4, 5}. (Mass function is zero at other values.) P {=x,+y =5} P {=x}p {Y =5 x} P { = x + Y = 5} = P {+Y =5} = P {+Y =5}. This is p(x)p(5 x) equal to 5. j=0 p(x)p(5 x) (b) The conditional expectation of Y given that = Y. ANSWER: Let y 0 be an integer. First we compute P {Y =y,=y} p(y)p(y) P {Y = y = Y } = P {=Y } =. Then we note k=0 p(k)p(k) that the E[Y = Y ] = P {Y = y = Y }y. y=0 (c) The probability mass function for Y given that > Y. ANSWER: Write Z = Y. Then 0 0 p Z (z) = P {Z = z} = P {Y = y}p { = z+y} = p(y)p(z+y). y= y=0 p Z (z) p Z (z) Now for z > 0 we have P {Z = z Z > 0} = P {Z>0} =. j= p Z (j) (d) The probability that = Y. ANSWER: k=0 p(k).. (5 points) Solve the following: (a) Let be a normal random variable with parameters (µ, σ ) and Y an exponential random variable with parameter λ. Write down the probability density function for + Y. ANSWER: f ((a y) µ) /σ +Y (a) = f (a y)f Y (y)dy = e λe λy dy 0 πσ (b) Compute the moment generating function and characteristic function for the uniform random variable on [0, 5]. ANSWER: Generally when is uniform on (a, b) we have tb e ta itb e M ita (t) = e and φ (t) = e t(b a) it(b a).

2 (c) Let,..., n be independent exponential random variables of parameter λ. Let Y be the second largest of the i. Compute the mean and variance of Y. ANSWER: This is essentially the radioactive decay problem: how long until n of the n particles have decayed. Let T k be time from when (k )th particle decays until kth particle decays. Each T k is exponential with parameter (n + k)λ. It has expectation (n+ k)λ and variance. By additivity of expectation (n+ k) λ n n E[ k= T k ] = k= E[T k ]. By independence of the T k we also have n n Var[ T k ] = Var[T k ]. 3. (0 points) k= k= (a) Suppose that the pair (, Y ) is uniformly distributed on the disc x + y. Find f, f Y. ANSWER: f(x, y) = π on the disc, zero elsewhere. Then for x (, ) we have f (x) = f(x, y)dy = π x. Symmetric argument gives f Y (y) = y for y (, ). π (b) Find also f +Y and f max(x,y). ANSWER: Easiest to first calculate F +Y (a) for a [0, ]. This is times the area of the disc of radius r with r π = a. Thus F +Y (a) = π πr = r = a. Conclude that + Y is uniform on [0, ] and f +Y = on [0, ]. I ll skip the f max(x,y) part since it s a bit messy but you could at least in principle compute F max(x,y) (a) = P {max(x, y) a} using some trigonometry (computing area of intersection of a quadrant with a circle) and then differentiate to get the density function. (c) Find the conditional probability density for given Y = y for y [, ]. ANSWER: Uniform on ( y, y ). (d) Compute E[ + Y ]. ANSWER:, by part (b)

3 4. (0 points) Suppose that i are independent random variables which n take the values and.5 each with probability /. Let = i= i. (a) Compute E. n ANSWER: By independence E[] = E[ i ] =.5 n. i= (b) Estimate the P { > 000} if n = 00. ANSWER: Let K be number of times i is, so that 00 K is number of times it is.5. Then = K.5 00 K = K / 00 K = K 00. Note that > 000 if and only if K 00 0, i.e., K 55. Now we have a standard binomial problem. What s probability to have at least 55 heads when we toss 00 coins. Standard deviation is npq = 5. So should be roughly Φ(). 5. (0 points) Suppose is an exponential random variable with parameter λ =, Y is an exponential random variable with λ =, and Z is an exponential random variable with parameter λ 3 = 3. Assume and Y and Z are independent and compute the following: (a) The probability density function f +Y ANSWER: λ (a y) λ f +Y (a) = f (a y)f Y (y)dy = λ e λ e y dy. (b) Cov(Y, + Y ) 0 ANSWER: E[Y ( + Y )] E[Y ]E[ + Y ] = E[ Y ] + E[Y ] E[Y ](E[] + E[Y ]) = E[ ]E[Y ] + E[]E[Y ] E[]E[Y ]E[] E[]E[Y ]E[Y ]. Recall that E[] = /λ and E[ ] = /λ and E[Y ] = /λ and E[Y ] = /λ. (c) E[max{, Y, Z}] ANSWER: Perhaps the easiest thing is to first compute F max{,y,z} (a) = P {max{, Y, Z} a} = F (a)f Y (z)f Z (a) = ( e λa )( e λa )( e λ3a ). Then recall the formula E[F max{,y,z} ] = F max{,y,z} (a)da. 0 3

4 (d) Var[min{, Y, Z}] ANSWER: Minimum is exponential with parameter λ + λ + λ 3 = 6. So variance is /6 = /36. (e) The correlation coefficient ρ(min{, Y, Z}, max{, Y, Z}). ANSWER: This one is a bit tricky. Idea is to argue first that min{, Y, Z} and max{, Y, Z} min{, Y, Z} are independent. Thus by bilinearity of covariance, we have Cov(min{, Y, Z}, max{, Y, Z}) = Varmin{, Y, Z} = /36, using the result of (d). From here we use the formula for ρ (though it requires input from (c), which we skipped). 6. (0 points) Suppose,..., 0 be independent standard normal random variables. For each i {, 3,..., 9} we say that i is a local maximum if i > i+ and i > i. Let N be the number of local maxima. Compute (a) The expectation of N. ANSWER: Each i {, 3,..., 9} has a /3 change to be a local maximum. (Basically, i has to be the largest among itself and two neighbors.) So E[N] = 8/3. (b) The variance of N. ANSWER: We need to compute E[N ]. Letting N i be if i is a local 9 maximum, zero otherwise, we have E[N ] = 9 i= j= E[N i N j ]. This sum includes (a) 8 terms of form E[N i N i ], which are each equal to /3. (b) 4 terms of form E[N i N i+ ] or E[N i N i ] which are each equal to zero (to neighbors can t both be local maxima). (c) terms of form E[N i N i+ ] or E[N i N i ]. Consider: five guys in a row, we need the probability that the second and fourth are both local maxima. Have /5 chance that second is largest among these five, and given that, have /3 chance that fourth is local maximum. So /5 chance in case second is largest, similarly /5 in case fourth is largest, so E[N i N i+ ] = /5 over all. (d) = 30 remaining terms. Sum all these up to get E[N ]. Then Var[N ] = E[N ] E[N]. 4

5 (c) The correlation coefficient ρ(n, ). ANSWER: Here ρ(n, ) = Cov(N, )/ Var(N)Var( ). Hard part remaining is to compute Cov(N, ). By bilinearity of expectation, this is 9 j= Cov(N j, ). These terms are all zero except when j =. So we just need to find Cov(N, ). Key step to compute E[N ]. If f is standard normal density, this can be written as f(x)f(y)f(z)x where the integral is taken over the portion of R for which y > x and y > z. I think I ll skip the part where we actually compute the integral. 7. (5 points) Give the name and an explicit formula for the density or mass function of n i= i when the i are (a) Independent normal with parameter µ, σ. ANSWER: normal with parameters (nµ, nσ ) (b) Independent exponential with parameter λ. ANSWER: gamma with parameters n and λ (c) Independent geometric with parameter p. ANSWER: negative binomial with parameter p. (d) Independent Poisson with parameter λ ANSWER: Poisson with parameter nλ. (e) Independent Bernoulli with parameter p. ANSWER: binomial with parameters (n, p). 5

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