0
$\begingroup$

This question already has an answer here:

If $X_1$, $X_2$, $\dots$, $X_k$ are random variables with Poisson distribution with the following parameters: $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_k$.

Show that $(X_1 \mid X_1 + X_2 + \dots + X_k)$ (conditional probability) has binomial distribution.

If I suppose that $(X_1 \mid X_1 + X_2 + \dots + X_k)$ has binomial distribution and given that the sum $Y = X_1 + X_2 + \dots + X_k $ has Poisson distribuition (with $\sum_i^k \lambda_i$ parameter), I can build the result and the distribution is a Binomial with parameters $(k, \frac{\lambda_1}{\lambda_1 + \lambda_2 + \dots + \lambda_k})$, but looks like I done from the end to beginning. There is a direct method?

I am asking for an explanation, and the similar post did not help me.

$\endgroup$

marked as duplicate by Mike Earnest, Clement C., Xander Henderson, Aweygan, Jose Arnaldo Bebita-Dris May 23 '18 at 9:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The reason this is a duplicate is because $X_2+\dots+X_k$ is also a Poisson random variable, with parameter $\lambda_2+\dots+\lambda_k$. $\endgroup$ – Mike Earnest May 22 '18 at 23:18
  • $\begingroup$ Is my answer right or wrong ? Could you tell ? $\endgroup$ – EduardoGM May 22 '18 at 23:41
  • 1
    $\begingroup$ You have the probability parameter correct, but the first parameter should be $X_1+X_2+\dots+X_k$, not $k$. $\endgroup$ – Mike Earnest May 22 '18 at 23:42
  • 2
    $\begingroup$ I got it ! For instance, $X_1+X_2+\dots+X_k=y$ so my binomial parameters are $n = y$ and $p$ that one written. Nice! $\endgroup$ – EduardoGM May 22 '18 at 23:49
1
$\begingroup$

In a Poisson process, point events arrive at a constant rate independent from occurance of prior events.

You have $k$ independent Poisson processes with rates $\lambda_1, \lambda_2,\ldots,\lambda_k$.   All in all, point events are arriving independently at total rate $\lambda_1+ \lambda_2+\ldots+\lambda_k$.   That is $\sum_{i=1}^k\lambda_i$.

One such event arrives!   Because (insert the justification), therefore the probability that it has done so in Process 1 is $\lambda_1/\sum_{i=1}^k\lambda_i$.   Call it a success if it is so.   This is a Bernoulli trial.

Many such events arrive!   The count of "successes" among these many independent Bernoulli trials, when given the total amount, will be Binomially distributed, by definition.

$$(X_1\mid (X_i)_{i=1}^k) \sim\mathcal{Binom}(\sum_{i=1}^k X_i, \lambda_1/\sum_{i=1}^k \lambda_i)$$

$\endgroup$
0
$\begingroup$

\begin{align} & \Pr(X_1 = x\mid X_1+\cdots+X_n=y) \\[10pt] = {} & \Pr(X_1=x\ \&\ X_1+\cdots+X_n=y)/\Pr(X_1+\cdots+X_n=y) \\[10pt] = {} & \Pr(X_1=x\ \&\ X_2+\cdots+X_n=y-x)/\Pr(X_1+\cdots+X_n=y) \\[10pt] = {} & \Pr(X_1=x)\cdot\Pr(X_2+\cdots+X_n=y-x)/\Pr(X_1+\cdots+X_n=y) \\[10pt] & \text{etc.} \end{align}

$\endgroup$
  • $\begingroup$ I can find several questions that are much simpler than mine and that are answered promptly, is it something related to my "reputation points"? $\endgroup$ – EduardoGM May 22 '18 at 23:51

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