Mgf @m(0) =1

By the property of MGF, M(0) = 1 always and this result can be used to crosscheck the MGF obtained for any PDF...

But for PDF, f(x) = 0.5(1-x), -1<x<1

the MGF is e^(-t) * 1/t^2 * {e^2t - 2t - 1}

But this MGF does'nt follow above rule

Pls tell me where i am wrong????

 

That looks like the correct MGF to me.

Problem is that it's not well defined at t=0, since
you have a 1/t^2 in there.

The limit of that MGF as t->0 is 1, I think.

 

but i core reading they have clearly written that for all MGFs M(t) must be 1 if t=0

 

I think you're right: M(0) = 1 for any random variable X that you care to think of, because (by definition):

M(t) = E[e^(tX)] => M(0) = E[e^0] = E(1) = 1.​

How did you find your expression for the m.g.f. - did you integrate by parts? If so, you might have started by writing something like:

M(t) = (1/2) * Integral(-1,1){e^(tx).(1-x) dx}​

Notice that it doesn't make sense to integrate the "e^(tx)" bit up to (1/t)e^(tx) when t = 0 (because you end up dividing by zero). You may have to consider the case t = 0 separately (like above).

 

You get an x*e^tx term to integrate and I don't see a way to integrate other than the standard "by parts" leaving a 1/t^2 term.

See the following site from wolfram

http://mathworld.wolfram.com/UniformDistribution.html

which has the MGF for a uniform distribution (Equation #10) which also would normally be undefined at t=0. However, no explanation is given for why it is set to 1.

Note that lower down it says you can find the limit as t->0 instead to generate moments where MGF is not defined.

Bit rusty myself in MGF and limits etc, but gut feeling is that limit of your MGF as t->0 is 1.

Just realised what MarkC was saying, ie since t=0 is not defined you set t=0 and evaluate the expression BEFORE any integration. This also makes sense and gives 1 directly (this is more or less proof that all MGF at t=0 are 1).