So I'm revising moment generating functions and I'm stuck on a part of a question I'm looking at.

So I am asked to find the moment generating function of a random variable X whose distribution is given by $$\Bbb{P}(X=1)=\Bbb{P}(X=-1)=1/2$$ I get the Moment generating function for this random variable X to be $$M(t)=\frac{e^t-e^{-t}}{2}$$I am then asked to show that $$M(t)\le \exp\left(\frac{t^2}{2}\right)$$ I am not sure how to go about doing this. I was thinking I could use Taylor series, and expand both sides of the inequality, but I seem to get stuck and can't actually show that one is less than or equal to the other.

Any help appreciated-thanks.

  • 2
    $\begingroup$ Shouldn't the moment generating function have the property that $M(0)=1?$ It should be $$M(t)=\frac{e^{t}+e^{-t}}{2}.$$ $\endgroup$ – Thomas Andrews Jun 6 at 19:39
  • $\begingroup$ You have $E(X^k)=1$ if $k$ is even, and $E(X^k)=0$ if $k$ is odd... $\endgroup$ – Thomas Andrews Jun 6 at 19:40
  • $\begingroup$ @Math1000 Actually, $$\frac{e^t+e^{-t}}{2}=\sum_{n=0}^{\infty}\frac{t^{2n}}{(2n)!}$$ (note the denominator $(2n)!$ rather than $n!.$ $\endgroup$ – Thomas Andrews Jun 6 at 19:48
  • $\begingroup$ You can also see Proving $\ln \cosh x\leq \frac{x^2}{2}$ for $x\in\mathbb{R}$. (Note that your MGF is $\cosh t$, so you want to show that $\cosh t \le e^{\frac{t^2}{2}}$, i.e. $\ln (\cosh t)\le \frac{t^2}{2}$.) $\endgroup$ – Minus One-Twelfth Jun 6 at 20:01

Your moment generating function is wrong. One way you can quickly tell: If $M$ is a moment-generating function, then $M(0)=1.$

It should be:

$$M(t)=\frac{e^t+e^{-t}}{2}=\sum_{k=0}^{\infty} \frac{t^{2k}}{(2k)!}$$

This is because $E\left(X^{2k}\right)=1$ and $E\left(X^{2k+1}\right)=0$ for integers $k\geq 0$.

The we have: $$\exp\left(\frac{t^2}{2}\right)=\sum_{k=0}^{\infty} \frac{t^{2k}}{2^k\cdot k!}$$

So, since $t^2\geq 0,$ if you can prove that $\frac{1}{(2k)!}\leq \frac{1}{2^k\cdot k!}$ for all $k,$ you'd be done. I'll leave that to you.


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