Conditional distribution of geometric variables

Question

Setting

Suppose X1 and X2 are independent with the common geometric distribution w(k; p). Determine the conditional distribution of X1 given that X1 + X2 = n.

Solution

My argument is

$$\Pr[X_1| X_1+X_2 = n] = \frac{\Pr[X_1 + X_2 = n| X_1 = x_1]\Pr[X_1 = x_1]}{\Pr[X_1+X_2=n]}$$

and

$$\Pr[X_1 = x_1] = q^{x_1} p$$

$$\Pr[X_1+X_2=n|X_1=x_1] = \Pr[X_2 = n - x_1] = q^{n-x_1}p$$ while $$\Pr[X_1+X_2=n] =\sum_{j=1}^n q^j p q^{n-j} p = q^n p^2$$

However, I am not confident with this solution since geometric variables should be memoryless? So I feel like I am missing something here.

Answer 1

Let $X_1$ and $X_2$ be i.i.d geometric random variables with parameter $q$ (and $p{}={}1-q$). Using the definition of conditional probabilities for discrete random variables, we have $$ \begin{eqnarray*} P\left( X_1{}={}x_1\,|\, X_1{}+{}X_2{}={}n\right)&{}={}&\frac{P\left( X_1{}={}x_1\,,\, X_1{}+{}X_2{}={}n\right)}{P\left( X_1{}+{}X_2{}={}n\right)}\newline &{}={}&\frac{P\left( X_1{}={}x_1\,,\, X_2{}={}n-x_1\right)}{P\left( X_1{}+{}X_2{}={}n\right)}\newline &{}={}&\frac{P\left( X_1{}={}x_1 \right)P\left( X_2{}={}n-x_1\right)}{\displaystyle \sum\limits_{k{}={}1}^{n-1} P\left( X_2{}={}n-k \,|\,X_1{}={}k\right)P\left(X_1{}={}k\right)}\newline &{}={}&\frac{P\left( X_1{}={}x_1 \right)P\left( X_2{}={}n-x_1\right)}{\displaystyle \sum\limits_{k{}={}1}^{n-1} P\left( X_2{}={}n-k \right)P\left(X_1{}={}k\right)}\newline &{}={}&\frac{p^{x_1 - 1}\,q\,p^{n-x_1 - 1}\,q}{\displaystyle \sum\limits_{k{}={}1}^{n-1} p^{k - 1}\,q\,p^{n-k - 1}\,q }\newline &{}={}&\frac{p^{n-2}\,q^2}{\displaystyle \sum\limits_{k{}={}1}^{n-1} p^{n-2}\,q^2 }\newline &{}={}&\frac{1}{\left(n-1\right)}\,. \end{eqnarray*} $$ This conditional distribution is independent of the parameter $q$.

Answer 2

Some things to note. $\operatorname{Pr}[X_1 = x_1] = q^{x_1 - 1}p$. You had forgotten the "$- 1$". Likewise for $\operatorname{Pr}[X_2 = n - x_2]$.

But, more importantly, the sum of geometric random variables gives negative binomial.

$$ \operatorname{Pr}[X_1 + X_2 = n] = \sum\limits_{\substack{i + j = n \\ i, j \in \{1, 2, \ldots\}}} \overbrace{q^{i - 1}p}^{\operatorname{Pr}[X_1 = i]}\underbrace{q^{j - 1}p}_{\operatorname{Pr}[X_2 = j]} = \binom{n - 1}{1}p^2q^{n - 2} $$

The binomial coefficient $\binom{n - 1}{1}$ is the number of solutions to $i + j = n$, where $i, j \in \{1, 2, \ldots\}$.