Let $X_1,...,X_k$ are independent random variables and each $X_i \thicksim Po(\lambda_i), \forall i \in\{1,...k\}$ then derive the conditional pdf of $X = (X_1,...X_k)$ in case $X_1 +...+X_k= n$.

Since two random variables case, such as $X_1 \thicksim Po(\lambda_1)$ and $X_2\thicksim Po(\lambda_2)$ which are independent and $X_1+X_2 = n $, $P(X_1=x_1\mid X_1+X_2 =n) = \begin{pmatrix}n\\x\end{pmatrix}\cdot\begin{pmatrix}\dfrac{\lambda_1}{\lambda_1+\lambda_2}\end{pmatrix}^{x_1}\cdot\begin{pmatrix}\dfrac{\lambda_2}{\lambda_1+\lambda_2}\end{pmatrix}^{x_2}$,

through the analogy of it, $k$ random variables case would be represented as following:

$P((X_1,...,X_k)=(x_1,...,x_k)\mid X_1+..+X_k =n)) \\= \begin{pmatrix}n\\x_1...x_k\end{pmatrix}\cdot\begin{pmatrix}\dfrac{\lambda_1}{\lambda_1+..+\lambda_k}\end{pmatrix}^{x_1}\cdot\cdot\cdot\begin{pmatrix}\dfrac{\lambda_k}{\lambda_1+..+\lambda_k}\end{pmatrix}^{x_k}$

Is this reasoning available and renders right answer?

  • 1
    $\begingroup$ Not PDF (probability density function) but PMF (probability mass function). $\endgroup$ – drhab Jul 8 '17 at 18:24

Your analogous expression should be of the form,

$$ P\left((X_1,..,X_{k-1} = (x_1,..,x_{k-1}) | \sum\limits_i^kX_i = n\right) $$ since you're conditioning the first $k-1$ RV's on the sum. We have,

\begin{align} P\left((X_1,..,X_{k-1} = (x_1,..,x_{k-1}) | \sum\limits_i^kX_i = n\right) = \frac{P\left((X_1,..,X_{k-1} = (x_1,..,x_{k-1}) , \sum\limits_i^kX_i = n\right)}{P\left(\sum\limits_i^kX_i = n\right)} \\ = \frac{n!}{x_1!x_2!\cdots(n-\sum_i^{k-1}x_i)!}\left(\frac{\lambda_k}{\lambda_1 +...+\lambda_k}\right)^n\left(\frac{\lambda_1}{\lambda_k}\right)^{x_1}\cdots\left(\frac{\lambda_1}{\lambda_k}\right)^{x_{k-1}}\end{align}

which after some algebra ($x_k = n - \sum\limits_i^{k-1}x_i$), seems to match your guess.

  • 2
    $\begingroup$ "Your analogous expression should be of the form..." Actually, considering $$P((X_1,...,X_k)=(x_1,...,x_k)\mid X_1+..+X_k =n)$$ is perfectly all right. One could even argue that the more symmetrical formulas this yields are to be preferred. $\endgroup$ – Did Jul 8 '17 at 18:20

If $x_1,\dots,x_k$ are nonnegative integers with $x_1+\cdots+x_k=n$ then:$$P(X_1=x_1,\dots,X_k=x_k\mid \sum_{i=1}^kX_i=n)=\frac{P(X_1=x_1,\dots,X_k=x_k)}{P(\sum_{i=1}^kX_i=n)}$$

Working this out you will indeed arrive at the expression mentioned in your question.


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