Let $X \sim Bin(n,p)$

Part A.

Show that the argument

"Then $X/n \sim Bin()$ with $E[X/n]=p$and $Var[X/n]=pq/n$"

is false.

My book says we can prove this result using a moment generating function. But to me it makes no intuitive sense: We have Bin-distributed RV $X$ and if we rescale it with $1/n$ of course it will be Bin-distributed. And we know from standard formulas that $E[aX]=aE[X]$ and $Var[aX] = a^2Var[X]$ so I think the argument is correct.

(For full disclosure, the argument we should falsify according to the book is "Let X be a binomial random variable with mean np and variance np(1−p). Show that the ratio X/n also has a binomial distribution with mean p and variance p(1−p)/n." But I think the argument is correct. )

Ok, now I get part A. Here is Part B. Prove it using the moment generating function of $X/n$.

Attempt: I get $$m_X(t)=E[\exp(tX/n)]=(m_X(t))^{1/n}$$

  • 2
    $\begingroup$ I binomially distributed random variable $X$ with $n=6$ has support $\{0,1,2,3,4,5,6\}$. But $X/6$ has support $\{0 ,\, 1/6,\, 2/6,\, 3/6,\, 4/6,\, 5/6,\, 1\}$. That is not the support of a binomial distribution. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 16 '15 at 22:12

Of course $\frac{X}{n}$ isn't binomial distributed for $n > 1$, since $P(\frac{X}{n} = \frac{1}{n}) \ne 0$ and the binomial distribution only assumes integer values.

You argument shows nothing. Just because a random variable has a mean and a variance doesn't mean it is binomial distributed.

  • $\begingroup$ Thank you, this solved it. But I forgot one part of the question: Prove it using the moment generating function of $X/n$ and I get $$m_X(t)=E[\exp(tX/n)]=(m_X(t))^{1/n}$$ $\endgroup$ – jacob Sep 17 '15 at 22:38
  • $\begingroup$ You can't just pull $1/n$ out of the expectation. The correct result is $m_X(t/n)$. $\endgroup$ – Dominik Sep 18 '15 at 7:28

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