Let $X$ be a random variable with mgf $M_X(t)=a+be^{2t}$.Given that mean of the random variable $X$ is $1.5$. Then find the expected value of the random variable $2^X.$
My attempt:
We know that $E(X^r)=\frac{d^r}{dt^r}M_X(t) \rvert_{t=0}$.
So, we have that $E(X) = 2b = 1.5$.
This means that $b=0.75$.
Also, we know that $M_X(0)=1.$ So, we get $a=0.25$.
I am now going to do some operation here. I want to know whether it is valid or not.
Now, set $t= \ln 2$. So, we have that $e^{(\ln 2) X} = \big( e^{\ln2 } \big)^X = 2^X.$ From the definition $M_X(t)=E(e^{tX})$, we get $M_X(\ln 2) = E(e^{\ln 2 X}) = E(2^X) = 0.25 + 0.75 e ^ {2 \times \ln 2} = 3.25$.
Hence, we have that $E(2^X)=3.25$.
Have I gone wrong somewhere ?
If yes, please let me know where I have gone wrong and what is the correct approach.
If not, then suggest what property of the random variables or distributions I am using here.
Please help.
Thank you very much.
