I've asked before about a Problem from Grimmet$Welsh (and I thank a lot to @angryavian and @Graham Kemp):

"If $X + Y$ and $X - Y$ are independent, show that \begin{align} M\left(2t\right) = M\left(t\right)^{3}M\left(-t\right), \end{align} where $X,Y$ are independet r.v. with mean $0$, variance $1$ and $M(t)$ finite."

This is it's link: Moment generating function applied in $2t$.

But now there's the "second" part of the problem: To show that $X$ (and $Y$) are a r.v. with normal distribution with mean $0$ and variance $1$.

The book itself sugest to define a function $\psi(t) = \frac{M(t)}{M(-t)}$ and show that $\psi(t)=\psi(2^{-n}t)^{2n}$. Then, show that $\psi(t) = 1 + o(t^{2})$ as $t \to 0$ and $\psi(1)=1$ when $n \to 0$. This will allow us to conclude that $M(t)=M(-t)$ and, when we apply this to the main equation (the one in the link and above) we get $M(t)=M(\frac{1}{2}t)^{4}$. The book then says to repeat the precedure to obtain the desired result. So, I have some questions:

  1. How to show that $\psi(t)=\psi(2^{-n}t)^{2n}$ ?
  2. What does that "o" in $\psi(t) = 1 + o(t^{2})$ means? (I can't remember of seeing this through the Chapter)
  3. What is the procedure to repeat in order to obtain the desired result? The entire one? The last part?

Of course, if somebody knows some another way to prove this statement, I'll be very aprecciated! Thanks in advance for your help!

  • 1
    $\begingroup$ Reference: Little-Oh notation: wolfram and Wiki $\endgroup$ Nov 25 '20 at 3:55
  • $\begingroup$ @Graham Kemp Interesting, I only knew the Big-O, using it to know the order of errors in Numerical Methods. $\endgroup$ Nov 25 '20 at 16:23


  • Using the first part of the question, $\psi(t) = \frac{M(t)}{M(-t)} = \frac{M(t/2)^3 M(-t/2)}{M(-t/2)^3 M(t/2)}$. Do some more work to show $\psi(t) = \psi(t/2)^2$.

  • $\psi(t) = \psi(t/2)^2$ implies the more general equality $\psi(t) = \psi(t/2^n)^{2n}$.

  • By a Taylor expansion of $\psi$, we have $\psi(t) = \psi(0) + \psi'(0) t + \frac{1}{2} \psi''(0) t^2 + \frac{1}{6} \psi'''(\xi)t^3$ for some $\xi$ between $0$ and $t$. We know $\psi(0)=1$. We have $$\psi'(t) = \frac{M'(t)M(-t) + M(t)M'(-t)}{M(-t)^2}$$ so $\psi'(0)=0$ (because $M'(t)=E[X]=0$). We also have $$\psi''(t) = \frac{d}{dt}\frac{M'(t)M(-t) + M(t)M'(-t)}{M(-t)^2} = \frac{[M''(t)M(-t) - M(t) M''(-t)]M(-t)^2 + [M'(t)M(-t)+M(t)M'(-t)] 2 M(-t) M'(-t)}{M(-t)^4}$$ so $\psi''(0)=0$ (since $M''(0)=E[X^2]=1$). So the Taylor expansion becomes $$\psi(t) = 1 + \frac{1}{6} \psi'''(\xi) t^3.$$ If you show $\psi'''$ is bounded by some constant $C$ for $t$ near zero (I can't think of a simple way to show this, and I am assuming third moments exist... maybe someone else can clean up my mess here), then we have $$\lim_{t \to 0} \frac{|\psi(t) - 1|}{t^2} = \lim_{t \to 0} \frac{C}{6} |t| \to 0$$ which is the definition of $\psi(t)=1+o(t^2)$.

  • $\psi(1) = \lim_{n \to \infty} \psi(2^{-n})^{2n} = \lim_{n \to \infty} (1 + o(2^{-2n}))^{2n} = 1$ (I am skipping steps here)

  • $\psi(1)=1$ implies $M(t)=M(-t)$

  • $M(2t) = M(t)^3 M(-t) = M(t)^4$

  • $M(t) = M(t/2)^4$

  • $M(t) = M(t/2^n)^{4n}$

  • some more work needs to be done to deduce that $M(t)=e^{-t^2/2}$ is the only possible candidate MGF that satisfies the above recurrence.

  • $\begingroup$ To show that $M(t)=e^{-\frac{t^2}{2}}$ is the only possible candidate, may I use the uniqueness theorem? Seems to fit nicely $\endgroup$ Nov 25 '20 at 16:17

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