Can You Derive the MGF of Y=-X Using MGF of X Directly?

[zli034]

Say r.v. X, we have pdf f(x) and mgf Mx(t) defined.

Then define Y=-X, y is negative x.

Can we get mgf of Y, i.e. My(t) and how?

I know I can go the way to get pdf f(y) first then My(t). I want to know if Mx(t) is already in my hands, it should be easier to get My(t) other than do f(y) first.

I thought about this whole afternoon. Just don't get it. Please help and thanks in advance.

xoxo

 

[statdad]

Remember that for any random variable

$$M_x(t) = E[e^{tx}]$$

If you want the mgf for cX (any constant times X)

$$M_{cX}(t) = E[e^{t(cx)}]$$

How can you simplify this expression, how does it relate to $$M_X(t)$$,
and how do both observations relate to your problem?

 

[zli034]

I have found one way.

I have here x=log(1-x) and f(x) also known.

So we can see if I put x=log(1-x) into Mcx(t), it can be simplified easily. Here is my solution for this problem and it depends on the form of function of x.

 

[statdad]

How does your second post relate to your first question?