Thread: Finding E[XY] in termes of the moment generating function

The random variable X has the N(μ,theta^2) distribution.

a) Show the moment generating fumction-have done this.

b) Define Y = e^−X and deduce from part (a) the values of E[Y ],
E[Y^2] and hence var(Y ).

c) Give an expression for E[XY ] in terms of MX(t) and hence calculate E[XY ].

d)Show that the covariance of X and Y can be written in the form
Cov(X, Y ) = -theta^2exp(theta^2/(2-μ) and derive an expression for the correlation.

have done part a.

Have done part b i think, since the moment generating function is
E[exp(tx)] let t=-1. then E[exp(-x)]=E[y]. Same foe t=-2 for E[y^2], then subtract for the variance.

Little stuck here and on the next section, we want E[xy], in terms of the moment generating function and then calculate.

The covariance is given by: E[xy]-E[x]E[y], hence from the part before this should make more sense, correlation, will be generated from what we have found previous,

Just really struggling on part c, the rest will follow from that,
any help would be most appreciated, many thanks.

Hello,

Well you have to think about it, there's an X in front of the exponential. That reminds of a derivative. So what happens if you differentiate with respect to t ? What will be the derivative of wrt to t ? (let's say you are allowed to take its derivative).

If you still don't see it, remember that

Spoiler:

the expectation is an integral...

Spoiler:

...so it's like derivating under an integral sign