Find the moment generating function for X if $f(x) = 2x$ , $0 < x < 1 $. Then, use the moment generating function to find $\mu_{x}$ and $\sigma_{x}^{2}$.
I calculated and found the value of MGF or moment generating function. Its $M.G.F. = [2e^{t}/t - 2e^{t}/t^{2} + 2/t^{2}]$ which is correct I know. How am I supposed to find the rest two?
I used a concept where I have $N(\mu,\sigma)$, and so you let $Z = (X - \sigma)/\mu$, then since our function is from 0 to 1, therefore we should be having $N(0,1)$ which means $\mu = 0$, and $\sigma = 1$. So, you get $Z = X$ which is basically the same thing. I got confused there. Also, I was thinking whether to find mean and variance of r.v. X. Can someone please help me on this?
Indeed, the moment generating function is \begin{align} M_X(t) &= \mathbb E[e^{tX}]\\ &= \int_{\mathbb R} e^{tx}f_X(x)\ \mathsf d x\\ &= \int_0^1 2xe^{tx}\ \mathsf dx\\ &=\frac 2{t^2}(1 - e^t (1 - t)) \end{align} We can compute the $n^{\mathrm{th}}$ moment of $X$ by $\mathbb E[X^n]=\lim_{t\to 0}M_X^{(n)}(t)$. For $n=1$, this is \begin{align} \mathbb E[X] &= \lim_{t\to 0}M_X'(t)\\ &=\lim_{t\to 0} \frac2{t^3}(t^2e^t -2 te^t +2 e^t-2)\\ &=\frac23, \end{align} and for $n=2$ this is \begin{align} \mathbb E[X^2] &= \lim_{t\to 0}M_X''(t)\\ &=\lim_{t\to 0} \frac 2{t^4}(t^3e^t -3t^2 e^t +6 t e^t-6 e^t+6)\\ &= \frac12. \end{align} Hence $$ \mathrm{Var}(X) = \mathbb E[X^2] - \mathbb E[X]^2 = \frac 12 - \left(\frac 23\right)^2 = \frac1{18}. $$