Moment Generating Functions

**jandrewross** · September 20th 2007, 05:57 PM

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!

**CaptainBlack** · September 21st 2007, 10:53 AM

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!

Start with the definition of the moment generating function:

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

then:

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

Now let $y=-x$ , then:

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

RonL

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