I have a parameter $\theta$ which is equal to $\sqrt{(X / n)}$, or the $\sqrt{\bar{X}}$.

$X$ follows a binomial distribution $\text{Bin}(n,\theta^2)$

I was trying to find the distribution using the M.G.F.. Is my approach to find the distribution of $\theta$ correct?

Let $p$ be $\theta^2:$

MGF of $X$ (binomial): $$[ (1-p) + pe^t]^n$$

MGF of $\bar{X}$: $$[ (1-p) + pe^{t/n}]^n$$

MGF of $\sqrt{\bar{X}}$: $$[ (1-p) + pe^{ \sqrt{t/n} } ]^n$$

I'm not really familiar with MGF's, your help will really be appreciated! :) Also, if anyone has a reference to a master collection of transformations with MGF's, it will really help as well! The one's I've found online deal with a very small subset of examples, eg addition.

Thank you!

  • 2
    $\begingroup$ MGF of square root of r.v. does not equal to MGF of square root of variable. $\endgroup$ – NCh Jun 5 '17 at 2:20
  • $\begingroup$ If $X$ follows the binomial distribution $\text{Bin}(n,\theta^2)$, then $\theta$ is not $\sqrt{X/n}$. $\sqrt{X/n}$ is a random variable, not a parameter. $\endgroup$ – Robert Israel Jun 5 '17 at 18:51

If $X \sim \text{Binomial}(n, \theta^2)$ and $Z = \sqrt{X/n}$, the MGF of $Z$ is

$$ \eqalign{\mathbb E[\exp(t Z)] &= \sum_{x=0}^n \mathbb P(X=x) e^{t \sqrt{x/n}}\cr &= \sum_{x=0}^n {n \choose x} \theta^{2x} (1-\theta^2)^{n-x} e^{t \sqrt{x/n}}} $$

  • $\begingroup$ Hey Robert, thanks for your reply! This means that the distribution of Z is also binomial right? How do you get the parameters from this form though? $\endgroup$ – Wboy Jun 5 '17 at 3:34
  • $\begingroup$ No, the distribution of $Z$ is not binomial. $\endgroup$ – Robert Israel Jun 5 '17 at 18:47

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