1. ## Moment Generating Functions

Ok... this one is taking me forever so I could use some help...

Let X be a random variable with pdf f(x) and mgf M(t). Suppose f is symmetric about 0, (f(-x) = f(x)). Show that M(-t) = M(t).

If someone could figure this out for me I would really appreciate it.. I would really just love to know where to start!

2. Originally Posted by jandrewross
Ok... this one is taking me forever so I could use some help...

Let X be a random variable with pdf f(x) and mgf M(t). Suppose f is symmetric about 0, (f(-x) = f(x)). Show that M(-t) = M(t).

If someone could figure this out for me I would really appreciate it.. I would really just love to know where to start!

$\displaystyle M_X(t)=E(e^{tX}) =\int_{-\infty}^{\infty} e^{tx} p(x) dx$

then:

$\displaystyle M_X(-t)=E(e^{-tX}) =\int_{-\infty}^{\infty} e^{-tx} p(x) dx$

Now let $\displaystyle y=-x$, then:

$\displaystyle M_X(-t)=E(e^{-tX}) =\int_{\infty}^{-\infty} e^{ty} p(-y) (-1)dy$$\displaystyle =\int_{-\infty}^{\infty} e^{ty} p(y) dy=M_X(t)$

RonL