- Let Z be the (2×1) vector of jointly standard normal random variables.
(a) From Z get a jointly normal random variables X1 and X2 such that the means and variances of X1 and X2 are given as µ1 = 0.5, µ2 = 1 σ2 1 = 1, and σ2 2 = 0.5, and the covariance is given as σ12 = 0.5.
I don't see where this problem is heading. What should this mean? Maybe should I make the pdf of the standard normal random variables and then use the transformation technique?
Best Answer
You could start with: $X=\left(\begin{array}{c} X_{1}\\ X_{2} \end{array}\right)=\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)\left(\begin{array}{c} Z_{1}\\ Z_{2} \end{array}\right)+\left(\begin{array}{c} \mu_{1}\\ \mu_{2} \end{array}\right)=\left(\begin{array}{c} aZ_{1}+bZ_{2}+\mu_{1}\\ cZ_{1}+dZ_{2}+\mu_{2} \end{array}\right)$ where $a,b,c,d$ are constants.
Then it is assured that $X_{1}$ and $X_{2}$ have a joint normal distribution with $\mathbb{E}X_{i}=\mu_{i}$ for $i=1,2$.
If $A$ denotes matrix $\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)$ then the covariance matrix of $X$ is $AA^{T}=\left(\begin{array}{cc} \sigma_{X_{1}}^{2} & \mathsf{Cov}\left(X_{1},X_{2}\right)\\ \mathsf{Cov}\left(X_{1},X_{2}\right) & \sigma_{X_{2}}^{2} \end{array}\right)$.
This provides conditions on $a,b,c,d$ and actually you are asked to find a tuple $(a,b,c,d)$ that satisfies these conditions.