Determine the probability distribution with T = the sum of X

Question

Hi I currently have a problem on solving a probability distribution my professor is teaching atm.

I was ashamed to ask him about it because he knows my affinity of calculus and linear algebra is quite solid and he assumed my statistics & probability carries the weight.

Though I have been stuck for the past 1 day about this concept (tried reading online and all) and based of the example he has given us so far. So to start of given the following geometric distribution is

\begin{equation*} \Pr(X=x) = \left\{ \begin{array}{rl} \theta(1-\theta)^x\ & \text{if } x = 0,1,2,.. \text{if } 0<\theta<1\\ 0 & \text{otherwise } \end{array} \right. \end{equation*}

So from his example we attempted when: \begin{equation*} \Pr(X=3) = \theta(1-\theta)^3\ \end{equation*}

or when

\begin{equation*} \Pr(X<4) = \theta(1-\theta)^1 + \theta(1-\theta)^2+\theta(1-\theta)^3\ \end{equation*}

Then he states what about the probability distribution of \begin{equation*} T=\sum_{i=1}^n X_i \end{equation*}

So I attempt to just plug in the distribution \begin{equation*} \Pr(X=T) = \theta(1-\theta)^{\sum_{i=1}^n X_i} \end{equation*}

Then from here I use a bit of manipulation \begin{equation*} \Pr(X=T) = \theta(1-\theta)^{n \bar{X}} \end{equation*}

but now I am stuck at here. Is there anything else I should do? As I feel this can't be right.

Answer 1

It is customary, in the case of self-study questions, to provide a hint on how to solve the problem. I will therefore do that but refrain from giving the full answer. If you are still stuck, there is a final hint at the end.

I assume the variables $X_i$ are mutually independent.

The variable $T$ represents the sum of the values of $n$ observations of $X_i$. Each of the $X_i$ variables, in turn, represents the number of failed Bernoulli trials occurred before a successful one takes place, right? So we can view $T$ as representing the number of failed Bernoulli trials before we observe $n$ successful ones.

The thing is that for a single value of $T$, the sequence of Bernoulli trials can be different. For instance, if $n=2,T=3$, we can have $$ X_1=0,X_2=3 $$ or $$ X_1=1,X_2=2 $$ etc.

If you take a closer look, you'll realize that this outcome corresponds to 5 Bernoulli trials; on one hand $n=2$, so we keep going until we observe 2 successful trials. On the other, $T=3$, which means that we observed 3 failed attempts, so 2+3=5 attemps in total. Following your notation, we know that success occurs with probability $\theta$ and failure with probability $1-\theta$, so we can start to get an idea of how the probability of $T$ will look like.

Hint: What is the joint probability of independent variables?

The thing is, the successful trials might have occurred at different points in time. Maybe the first attempt was a success, and then we saw 3 failed trials in a row. Or perhaps we saw 2 failures, 1 success, 1 failure and 1 success.

So out of the five trials, we must pick the position of the two successful ones. The position of the failed trials will be determined after that. How many ways can we pick two positions out of five possible ones?

Hint: binomial coefficient.

There is one more thing to take into account, though. If we do trials until we observe $n$ successful ones, that means the last attempt was a success. Therefore, we only need to pick the position of $n-1$ successful trials out of $n+T-1$ possible choices.

Finally, the possible orderings (that is, positions of the successes) are disjoint outcomes, so we can simply sum their probabilities to get the probability of the union.

Hint: If you are still stuck, take a look at the negative binomial distribution.