Consider the PDF
$$f(x) = 3x^2, 0≤ x ≤ 1$$
I'm trying to get the first and third moment from this PDF. I found the $E(X)$ using the usual formula to be $3/4$. However, I found the MGF to be $$\varphi(t)=E(e^{tx})=\frac{({3t^2-6t+6})e^t-6}{t^{3}}$$
To find the first moment, I differentiated and obtained $$\varphi'(t)=\frac{({3t^3-9t^2+18t-18})e^t+18}{t^{4}}$$
But when substituting $t = 0$, I got $\varphi'(0)=0$ and so $E(X)=0$ which does not agree with what I had above. Any ideas will be appreciated. Thank you
You have used the formula that $E(X)=\varphi'(0)$. This is true only if $\varphi'$ is defined at $t=0$. In this case, $\varphi'$ is not defined at $t=0$. If you were to substitute $t=0$, you'd obtain $0/0$. Thus, one needs to take the limit $$E(X)=\lim_{t\to 0} \varphi'(t)=\frac{3}{4}$$ which agrees with what you had.