A pdf of random variable X is f(x)=1/2, support of X=[-1,1]
I am supposed to verify Mx(0)=1. How do we verify this? I found mgf of X: (e^t - e^-t)/2t
Also, when you take the first derviate of mgf, it should give you expected value of X based on definition but I am not getting that either. Expected value of X is 0. I couldn't verify this with mgf derivative.Please help!
Let $M_X$ be the MGF of $X$. Then, by definition, $$ M_X(0)=E[\exp(0\times X)]=E[\exp(0)]=E(1)=1 $$ As for $E(X)=0$: $$ E(X)=\int_{-1}^1xf(x)dx=\int_{-1}^1x/2dx=\int_0^1\frac{x}{2}+\int_{-1}^0\frac{x}{2}dx. $$ By a change of variable $u=-x$, we have $$ \int_{-1}^0\frac{x}{2}dx=\int_{1}^0\frac{-u}{2}(-du)=\int_1^0\frac{u}{2}du=-\int_0^1\frac{u}{2}du=-\int_0^1\frac{x}{2}dx. $$ Do you see why $E(X)=0$?