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If $m(t)$ is the moment generating function of a random variable, then so is $\frac12+\frac13 m(t)+\frac16 m(t)^2$. Explain why this is true.

This is for a 200 level course the proof cant be anything with later levels of stats. Any help would be greatly appreciated.

I have shown that since $m(t)$ is a mgf $m(0)=1$ and used that to show the mgf I'm trying to prove is its equivalence evaluated at $0$ is $1$ proving it exists. Found a mean and variance for the mgf further proving it exists. My prof says this is not the right way to go about it and I can not think of another way.

Also I know that the a mgf squared is the addition of random variables. I can only think of ways that the mgf exists and not how to prove the two are equivalent.

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    $\begingroup$ What have you tried? What's the definition of the mgf? What does its square look like? $\endgroup$
    – Glen_b
    Commented Oct 19, 2016 at 6:04
  • $\begingroup$ I have shown that since m(t) is a mgf m(0)=1 and used that to show the mgf im trying to prove is its equivalence evaluated at 0 is 1 proving it exists. Found a mean and variance for the mgf further proving it exists. My prof says this is not the right way to go about it and i can not think of another way $\endgroup$
    – tanner
    Commented Oct 19, 2016 at 6:16
  • $\begingroup$ I can only think of ways to prove that the m(t) exists but dont know where to start on showing the two are equal $\endgroup$
    – tanner
    Commented Oct 19, 2016 at 6:22
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    $\begingroup$ Unless I misread something, you won't be showing the two are equal, you'll be showing that the expression in $m$ is also an mgf of some random variable, not of the same random variable. $\endgroup$
    – Glen_b
    Commented Oct 19, 2016 at 7:27

1 Answer 1

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Big hint:

Consider this scenario, which will hopefully get you thinking along the right lines.

What would the MGF of a 50-50 mixture of $X$ and a degenerate distribution at $0$ look like?

i.e. imagine if you have some r.v. $X$ with mgf $m(t)$ and you define a new variable $Y= 0$ with probability $\frac12$ and $Y=X$ with probability $\frac12$. What does the MGF of $Y$ look like?

Does the result of that give you any ideas?

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  • $\begingroup$ o ok i think im starting to get it so i defining mine in terms of random variables i could say that X is a r.v with mgf m(t), and Y =0 with probability 1/2. $\endgroup$
    – tanner
    Commented Oct 19, 2016 at 14:33
  • $\begingroup$ So the mgf of Y could be m(t)= 1/2 + 1/2m(t), now if i look at my question I could say Y is a r.v with prob with 1/3, say X is a mgf with m(t), and I could define Z=X+Y so the mgf would look like mz(t)=1/2 + 1/3m(t) +(1/2)(1/3)mx+y(t) which i could show to be mz(t) = 1/2 +1/3m(t) + (1/6)m(t) ^ 2 . $\endgroup$
    – tanner
    Commented Oct 19, 2016 at 14:40
  • $\begingroup$ I think you made that slightly harder than it needs to be (though I guess it depends on your point of view) but something along the lines of what you were doing should work too. But do it more formally. $\endgroup$
    – Glen_b
    Commented Oct 19, 2016 at 14:44

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