I have a problem that I know I am very close to the solution for, but I think I just need some more formatting to make it a really clean proof.

The problem goes like this:

Suppose X is a discrete r.v. with the pmf p(x) = $(1-\theta)^{(x-1)}*\theta, x=1,2,3,...,$ and $0\le\theta\le1.$ Consider the statistic $T=\Sigma X_i$

i) Use the definition of a sufficient statistic to show that T is a sufficient statistic for theta.

Looking at this pmf, (and a hint from my professor saying," Does the pmf look familiar? It should."), I saw that this is a geometric distribution.

So going by the definition of sufficiency: $\frac{(P(X_1=x_1)P(X_2=x_2)***P(X_1=x_1)}{P(T=t)}$=H

Where H does not contain $\theta$.

For the top half of my equation, I get:

$\theta^n(\theta-1)^{(\Sigma X_i-1)}$

And for the denominator, I get:


So, I end up with:

$\frac{\theta^n(\theta-1)^{(\Sigma X_i-1)}}{\theta(\theta-1)^{(T-1)}}$

I know this does not cancel out to where my statistic is not dependent upon $\theta$.

I am not sure where I am going wrong with this one, but if someone could let me know, that would be greatly appreciated.

b. The other half is to prove this by factorization:

I get my factorization down to:


I am having the same problem as I am with a problem I posted earlier about a Beta distribution. I have g(T,$\theta$), but what would be my h? I know that the sum of geometric r.v.'s add up to one, so is my answer one again? :S


the ratio you end up with is an incomplete statement of Bayes' theorem. Actually, it must be: $$\frac{P(X=x)P(T=t \mid X=x)}{P(T=t)}.$$

But now you see that $P(T=t \mid X=x)$ works like an indicator function taking as value $1$ in the case x is such that $T(x)=t$, and $0$ otherwise.

In the first case you turn to the ratio you have derived and check that you can simplify the expression making it free from theta. In the other one the probability is just zero.

In both cases you see that the distribution of $X$ does not depend on $\theta$ and now you're done.


You have $$\Pr(X=x)= \theta (1-\theta)^{(x-1)}$$ and so with a sample of size $n$ you have $$\displaystyle \prod_{i=1}^n \Pr(X_i=x_i)= \theta^n (1-\theta)^{\sum_i (x_i-1)},$$ meaning that the sufficient statistics are $n$ and $\displaystyle \sum_{i=1}^n (x_i-1)$. But since $\displaystyle \sum_{i=1}^n (x_i-1) =\sum_{i=1}^n (x_i) - n$, you could also say $n$ and $\displaystyle \sum_{i=1}^n x_i$ are sufficient statistics.

The related probability mass function is the distribution of a binomial random variable with probability of success $\theta$ with $\displaystyle \sum_{i=1}^n x_i$ attempts and $n$ successes.

You may need to adjust this slightly if your geometric random variable can take the value $0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.