This is a multi-part question from Michael Steele's book.a) Let V be a multivariate Gaussian column vector with mean and covariate , and let A be an n x n real matrix. Show that AV is a multivariate Gaussian vector with mean and covariance . b) Show that if X and Y are independent Gaussians with mean zero and variance one, then X-Y and X+Y are independent Gaussians with mean zero and variance two.c) Prove that if X and Y are jointly Gaussian and Cov(X,Y)=0 then X and Y are independentd) Prove that if (X,Y) are jointly Gaussian, then the conditional distribution of Y given that X=x is normal with mean:and variance

For the first questionEdit: sorry for the notation I've changed

QuoteOriginally posted by: will0t5This is a multi-part question from Michael Steele's book.This is a good exercise which probably belongs in the student forum, but see also below.Quotea) Let V be a multivariate Gaussian column vector with mean and covariate , and let A be an n x n real matrix. Show that AV is a multivariate Gaussian vector with mean and covariance . E[AV]=A mu and Var[AV]=A Sigma A' are true for any multivariate random variable V, with mu = E[V] and Sigma = Var[V]. To show that AV is Gaussian, we can use the characteristic function: E[exp(i omega' AV)]=E[exp(i ( A' omega )' V)]=exp( i mu'(A' omega) - 1/2 * (A' omega)' Sigma (A' omega) )=exp( i (A mu)'omega - 1/2 * omega' (A Sigma A') omega ), proves that AV is a multivariate Gaussian.Quoteb) Show that if X and Y are independent Gaussians with mean zero and variance one, then X-Y and X+Y are independent Gaussians with mean zero and variance two.Applying (a) with A = [1 -1] and A = [1 1], respectively, shows that X-Y and X+Y are Gaussian with mean 0 and variance 2.For independence, by linearity, Cov[X-Y,X+Y]=E[X^2]-E[Y^2]=0, and then we can use part (c) below.Quotec) Prove that if X and Y are jointly Gaussian and Cov(X,Y)=0 then X and Y are independentThis can be seen immediately by looking at the form of the pdf of a multivariate normal. Is there another way?Quoted) Prove that if (X,Y) are jointly Gaussian, then the conditional distribution of Y given that X=x is normal with mean:and varianceI have not tried direct calculation. Does someone know a nice way to prove (d)? I have been asked questions in interviews that can be answered immediately using (d), but apparently there is a "clever" way of doing such questions.

I have an answer I think for the last question but I don't think it's a clever method. In the bivariate normal distribution you can express the pdf with with so z can be reexpressed as so you can decomposed easily the conditionnal distribution (left term) to the marginal one (right side). I think it should work that way.

just do "de-correlation", i.e., find a linear combination of X and Y: Z=Y+a*X which is uncorrelated with X, or Cov(Y+a*X,X)=0, thus a=-Cov(X,Y)/Var(X).Z is Gaussian with mean of mu_Y+a*mu_X=mu_Y-Cov(X,Y)/Var(X)*mu_X, and variance of Var(Y)+2*a*Cov(X,Y)+a^2*Var(X)=Var(Y)-[Cov(X,Y)]^2/Var(X).since Z and X are independent, so conditional on X=x does not change all the quantities above, hence E[Y|X=x]=E[Z-a*x|X=x]=E[Z]-a*x and Var[Y|X=x]=Var(Z)

QuoteOriginally posted by: wileyswjust do "de-correlation", i.e., find a linear combination of X and Y: Z=Y+a*X which is uncorrelated with X ... Yes, that was surely the argument they were looking for. Thanks.

Quotec) Prove that if X and Y are jointly Gaussian and Cov(X,Y)=0 then X and Y are independentThis can be seen immediately by looking at the form of the pdf of a multivariate normal. Is there another way?Is this a sufficient answer??Cov(X,Y) = E[XY] - E[X]E[Y] = 0 Therefore E[XY] = E[X] E[Y] meaning they are independent.

vxs, uncorrelated variables are not necessarily independent. in another thread, eugwigner has pointed to this example