• We are pleased to announce that the winner of our Feedback Prize Draw for the Winter 2024-25 session and winning £150 of gift vouchers is Zhao Liang Tay. Congratulations to Zhao Liang. If you fancy winning £150 worth of gift vouchers (from a major UK store) for the Summer 2025 exam sitting for just a few minutes of your time throughout the session, please see our website at https://www.acted.co.uk/further-info.html?pat=feedback#feedback-prize for more information on how you can make sure your name is included in the draw at the end of the session.
  • Please be advised that the SP1, SP5 and SP7 X1 deadline is the 14th July and not the 17th June as first stated. Please accept out apologies for any confusion caused.

Mgf @m(0) =1

J

jatin.agrawal

Member
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.
 
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:rolleyes:
 
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).
 
Last edited by a moderator:
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).
 
Back
Top