Understanding the Problem: Geometric Random Variables

The problem involves two independent random variables, $X_1$ and $X_2$, each following a geometric distribution. The probability mass function (PMF) for a geometric distribution starting at $k=0$ is given by $P(X=k) = q^k p$ for $k = 0, 1, 2, \dots$, where $p$ is the probability of success and $q = 1-p$ is the probability of failure.

We are asked to find the conditional distribution of $X_1$ given the sum $X_1 + X_2$. Let $Y = X_1 + X_2$. We want to find the distribution of $X_1$ given $Y=n$, for a fixed value $n \ge 0$. This is represented by the conditional probability $P(X_1=k | X_1+X_2=n)$.

Calculating the Conditional Probability

Using the definition of conditional probability, we have:

$$ P(X_1=k | X_1+X_2=n) = \frac{P(X_1=k \text{ and } X_1+X_2=n)}{P(X_1+X_2=n)} $$

The event "$X_1=k$ and $X_1+X_2=n$" is the same as the event "$X_1=k$ and $X_2=n-k$". This can only happen if $k$ is a possible value for $X_1$ and $n-k$ is a possible value for $X_2$. Since both $X_1$ and $X_2$ can take any non-negative integer value starting from 0, this means $k \ge 0$ and $n-k \ge 0$, which simplifies to $0 \le k \le n$.

Because $X_1$ and $X_2$ are independent, the probability of the joint event "$X_1=k$ and $X_2=n-k$" is the product of their individual probabilities:

$$ P(X_1=k \text{ and } X_2=n-k) = P(X_1=k) P(X_2=n-k) $$

Substituting the geometric PMF:

$$ P(X_1=k \text{ and } X_2=n-k) = (q^k p) (q^{n-k} p) = p^2 q^{k + (n-k)} = p^2 q^n $$

This probability is valid for $k=0, 1, \dots, n$.

Distribution of the Sum $X_1 + X_2$

The sum of two independent and identically distributed geometric random variables (starting from 0) follows a Negative Binomial distribution. Specifically, $Y = X_1 + X_2$ follows a Negative Binomial distribution with parameters $r=2$ (number of successes) and $p$ (probability of success on each trial). The PMF for this form of Negative Binomial distribution (number of failures before the 2nd success) is:

$$ P(Y=n) = \binom{n+r-1}{r-1} p^r q^n = \binom{n+2-1}{2-1} p^2 q^n = \binom{n+1}{1} p^2 q^n = (n+1) p^2 q^n $$

for $n = 0, 1, 2, \dots$.

Deriving the Conditional Distribution

Now we can substitute the probabilities we found back into the conditional probability formula:

$$ P(X_1=k | X_1+X_2=n) = \frac{P(X_1=k \text{ and } X_1+X_2=n)}{P(X_1+X_2=n)} = \frac{p^2 q^n}{(n+1) p^2 q^n} $$

Assuming $p \ne 0$ and $q \ne 0$, we can cancel the $p^2 q^n$ terms:

$$ P(X_1=k | X_1+X_2=n) = \frac{1}{n+1} $$

This result is valid for $k = 0, 1, \dots, n$.

Identifying the Conditional Distribution Type

The conditional probability $P(X_1=k | X_1+X_2=n)$ is constant for all possible values of $k$ within the range $0 \le k \le n$. The possible values for $X_1$ given that $X_1+X_2=n$ are $0, 1, 2, \dots, n$. There are $(n-0+1) = n+1$ such values. Since each of these $n+1$ values has the same probability $\frac{1}{n+1}$, this is the definition of a discrete uniform distribution over the set of values $\{0, 1, \dots, n\}$.

Therefore, the conditional distribution of $X_1$ given $X_1 + X_2 = n$ is a Uniform distribution.