MGF and characteristic function of a sum of iid random variables

Question

Suppose $X_1,X_2,...X_n$ are iid random variables with the density $$ f(x)=2x ,\,0 \leq x \leq 1.$$ Define $$Y= X_1+X_2+...+X_n.$$One way to find the expected value of $Y$ is to first find the mgf of Y using $$M_{X_1+X_2+...+X_n}(t)=\prod_{i=0}^{n}M_{X_i}(t)$$ where $$ M_{X_i}(t)=E(e^{t X_i})= \int_0^12xe^{tx}dx=\frac{2}{t^2}(te^t-e^t+1) $$ so that $$M_Y(t)=\left[\frac{2}{t^2}(te^t-e^t+1)\right]^n .$$ How do we calculate $E(Y)$ from here?I suspect $\frac{d}{dt}M_Y(t)|$ at ${t=0}$ does not exist.Further ,is it possible to find the pdf of $Y$?Thanking you in advance for any responces.

Answer 1

Computing the Taylor series of your MGF, we have \begin{align*} e^t&=1+t+\frac{t^2}{2}+\frac{t^3}{6}+\dots \\ te^t+1-e^t&=\left(t+t^2+\frac{t^3}{2}+\frac{t^4}{6}+\dots\right)-t-\frac{t^2}{2}-\frac{t^3}{6}+\dots \\ &=\frac{t^2}{2}+\frac{t^3}{3}+\dots \\ \frac{2}{t^2}(te^t+1-e^t)&=1+\frac{2t}{3}+\dots \\ \left(\frac{2}{t^2}(te^t+1-e^t)\right)^n&=1+\frac{2nt}{3}+\dots \end{align*} and so $$\mathbb{E}[Y]=\frac{2}{3}\cdot n$$

An easier way to compute $\mathbb{E}[Y]$ would have been to use linearity of expectation, which implies $$\mathbb{E}[Y]=\sum_j{\mathbb{E}[X_j]}$$

Computing the p.d.f. is possible, but hard. In general, $$\mathrm{pdf}(A+B)(t)=\int_{\mathbb{R}}{\mathrm{pdf}(A)(s)\mathrm{pdf}(B)(t-s)\,ds}$$ So one could go compute the p.d.f.s for the sum of the first few terms, conjecture an inductive formula, and then prove it. Alternatively, the pdf is the inverse bilateral Laplace transform of the MGF, and so (if you know complex analysis) $$\mathrm{pdf}(Y)(y)=\frac{1}{2\pi i}\int_{r-\infty i}^{r+\infty i}{\left(\frac{2}{t^2}(te^t+1-e^t)\right)^ne^{ty}\,dt}$$ where $r$ is any nonzero real.