Page 1 of 7� � Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) Solutions to Quiz 2: Spring 2006 Problem 1: (30 points) Each of the following statements is either True or False. There will be no partial credit given f or the True False questions, thus any explanations will not be graded. Please clearly indicate True or False in the below, ambiguous marks will receive zero credit. All parts have equal weight. Points breakdown: Each question carries 3 points (a) X and Y are independent random variables. X is uniformly distributed on the interval [−2, 2], while Y is uniformly distributed on the interval [−1, 5]. If Z = X + Y , then fZ(3) = 1/6. True Since Z = X + Y and X,Y are independent, the PDF of Z (fZ(z)) can be obtained by convolving the PDFs of X (fX(x)) and Y (fY (y)). That is, fZ(z) = fX(u)fY (z − u) du = fY (u)fX(z − u) du. However, since we are interested in only fZ(3), we need to evaluate the convolution integral at only one point (z = 3). T he convolution sum and the corresponding figure are sh own below. � u=5 � u=5 1 1 1 1 fZ(3) = fY (u)fX(3 − u) du = ( )( ) du = 4 = u=1 u=1 6 4 24 6 1 u 1/6 1/4 −1 0 3 fz(u) fx(z−u) 5 (z−2) (z+2) Figure 1: fY (u) is uniform between −1 and 5. fX(z − u) is uniform between z − 2 and z + 2. (b) If X is a Gaussian random variable with zero mean and variance equal to 1, then the density function of Z = |X| is equal to 2fX(z), z ≥ 0. True Here Z is a derived random variable defined by Z = |X|. We can obtain the PDF of Z using the standard technique of finding the CDF of Z and then obtaining th e PDF by differentiating the CDF. We have (for z ≥ 0) FZ(z) = P(Z ≤ z) = P(|X| ≤ z) = P(−z ≤ X ≤ z) = FX(z) − FX(−z).Page 2 of 7Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) Taking derivatives, fZ(z) = fX(z) + fX(−z) = 2fX(z) since the standard Gaussian PDF is symmetric (c) The sum of a random number of independent Gaussian random variables with zero mean and unit variance results in a Gaussian random variable regardless of the distribution of N (the number of sums). False This can be verified by taking the transform of the new random variable. If X1, X2 . . . XN are IID Gaussian random variables and N is also a random variable in dependent of the Xis, then the transform of the sum Y = X1 + X2 . . . XN is given by MY (s) = MN(s)|es=MX(s). 2 2 s σThis does not take the form esµe 2 for all MN(s). (d) If X and Y are independent r andom variables, both exponentially distributed with parameters λ1 and λ2 respectively. Then the random variable Z = min{X, Y } is also exponentially distributed. True Here Z is a derived random variable defined as Z = min{X, Y }. We can obtain the PDF of Z by first determining its CDF and then taking the derivative. The CDF of Z is given by FZ(z) = P(Z ≤ z) = P(min{X, Y } ≤ z). It is not very straight forward to determine this probability. Instead, we can easily obtain P(Z ≥ z). Since this is equ ivalent to 1 − FZ(z), we have 1 − FZ(z) = P(Z > z) = P(min{X, Y } > z) = P(X > z, Y > z) = P(X > z)P (Y > z) since X,Y are independent = exp(−λ1z) exp(−λ2z) = exp(−(λ1 + λ2)z) Taking the derivative, we have fZ(z) = (λ1 + λ2) exp(−(λ1 + λ2)z). This is the pdf of an exponential random variable with parameter (λ1 + λ2). (e) L et the transform associated with a random variable X be � s �15 eMX(s) = . 1 − sPage 3 of 7� � � � Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) Then E[X] is equal to 30. True A straight forward way to confirm this f act is to compute the expected value by taking the derivative of MX(s) and then evaluating it at s = 0. We have dMX(s) es �15 (1 − s)es − es(−1) � = 15 ds 1 − s (1 − s)2 dMX(s)� E[X] = � = 15 · 2 = 30 ds s=0 The next set of questions are concerned with two independent rand om variables: Y is normal with mean 0 and variance 1, and X is uniform between [0, 1]. Z = X + Y . (f) The conditional density of Z given X, fZ|X(z|x), is normal with mean x and variance 1. True Given the value of x, the random variable Z is a derived random variable given by Z = x +Y . This is a normal random variable with mean x + E[Y ] and variance var(Y ). (g) var(Z) = 2. False var(Z) = var(X) + var(Y ) since X and Y are independent 1 = + 1 �= 2 12 (h) E[X | Z = −1] = − 1. False Since X is u niformly distributed between 0 and 1, the expected value cannot take on negative values. (i) cov(X, Z) = var(X) True By definition, we have cov(X, Z) = E[XZ] − E[X]E[Z]. Using Z = X + Y , we have cov(X, Z) = E[X(X + Y )] − E[X]E[X + Y ] = E[X2] + E[XY ] − E[X](E[X] + E[Y ]) = E[X2] − (E[X])2 + E[XY ] − E[X]E[Y ] = var(X) + cov(X, Y ) Since X and Y are independent, cov(X, Y ) = 0. (j) Z = E[X | Z] + E[Y | Z] True Since Z = X + Y , we have E[Z|Z] = E[X + Y |Z] conditional expectation is linear Z = E[X|Z] + E[Y |Z] since E[Z|Z] = ZPage 4 of 7� � � � Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) Problem 2: (25 points) Points breakdown: (a) 7 points; (b)–(d) 6 points each The continuous random variables X and Y have a joint pdf given by y x 2 1 1 2 c, if (x, y) belongs to the shaded region; fX,Y (x, y) = 0, otherwise. In class we have sh own the min imum least squares estimate of Y is given
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