Find Moment Generating Function from a non linear function Let $X$ be a r.v which has the following $mgf$ $$M_{x}(t) = (1-2t)^{-2}$$
If $Y = (100)(1.1)^{X}$ then calculate $E[Y]$
I know that $X \sim \chi^{2}_{(4)}$ but I am supposed to answer it only with $M_{x}(t)$
I think I have to start like this $E[e^{tY}] = E[e^{t(100)(1.1)^{X}}]$ but since the relation between Y and X is not linear I really don't know how to proceed.
Can anyone give me any hint? Or is it possible I can't solve it only with $M_{x}(t)$
 A: We want to find $E[Y]$ using only the moment generating function provided for X.
Step 1
We use the definition $Y = (100)(1.1)^X$, and substitute this into the above so that we are now looking to find the value $E[Y] = E[(100)(1.1)^X]$.
Step 2
We know that we can always factor out constants from the expectation and so we are able to rewrite the desired quantity as $100E[(1.1)^X]$
Step 3
You correctly observed that the function of X in the expectation is not linear, and so we are able to use a trick that is often useful in mathematics (particularly when it comes to probability and statistics), that $a^b=e^{ln(a^b)}=e^{(b)(ln(a))}$.
And so when we apply this trick to the desired quantity, we note that $100E[(1.1)^X] = 100E[e^{(X)(ln(1.1))}]$
Step 4
We notice that this is now in the form of the moment generating function which are given in the question: $M_x(t)=E[e^{tX}]=(1-2t)^{-2}$.
Therefore, $100E[e^{(X)(ln(1.1))}]=100M_x(ln(1.1))=100(1-2ln(1.1))^{-2}=152.65$.
And so we conclude that $E[Y] = 152.65$ using only the moment generating function provided for X.
