Question

Let \(X\) and \(Y\) be independent normal random variables, each having parameters \(\mu\) and \(\sigma^{2}\). Show that \(X+Y\) is independent of \(X-Y\). Hint: Find their joint moment generating function.

Short answer

In order to prove that X+Y is independent of X-Y, we found the joint moment generating function \(M_{X+Y, X-Y}(s,t)\) and showed that it can be factored into the product of the individual MGFs of X+Y and X-Y. This indicates that the two variables are indeed independent of each other.

Step-by-step solution

Find MGF of X and Y individually

Since X and Y are independent and have the same parameters, their moment generating functions (MGFs) are the same. Recall that the MGF of a normal distribution with mean μ and variance σ^2 is given by: \(M_X(t) = M_Y(t) = e^{\mu t + \frac{1}{2}\sigma^2t^2}\)

Find the joint MGF of X+Y and X-Y

To find the joint MGF of X+Y and X-Y, we need to find the expectation of the product of the exponentials of the variables multiplied by their corresponding t parameters. Since X and Y are independent, their joint MGF is given by: \(M_{X+Y, X-Y}(s,t) = E[e^{s(X+Y) + t(X-Y)}]\)

Simplify the joint MGF

Next, we need to simplify the expression and split it into the product of separate expectations, using properties of exponents: \(M_{X+Y, X-Y}(s,t) = E[e^{sX\hspace{0.5mm}+\hspace{0.5mm}tX\hspace{0.5mm}+\hspace{0.5mm}sY\hspace{0.5mm}-\hspace{0.5mm}tY}]\) \(M_{X+Y, X-Y}(s,t) = E[e^{(s+t)X}\hspace{0.5mm} e^{(s-t)Y}]\)

Separate expectations and apply individual MGFs

Since X and Y are independent, we can separate their expectations and apply the individual MGFs of X and Y: \(M_{X+Y, X-Y}(s,t) = E[e^{(s+t)X}] \hspace{0.5mm} E[e^{(s-t)Y}]\) By substituting the MGFs of X and Y, we get: \(M_{X+Y, X-Y}(s,t) = e^{\mu(s+t) + \frac{1}{2}\sigma^2(s+t)^2} \hspace{0.5mm} e^{\mu(s-t) + \frac{1}{2}\sigma^2(s-t)^2}\)

Proving independence of X+Y and X-Y

Since the joint MGF of X+Y and X-Y can be factored into the product of the individual MGFs, X+Y and X-Y are independent. Thus, we have shown that if X and Y are independent normal random variables with the same parameters, their sum (X+Y) is independent of their difference (X-Y).