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Suppose $Y_1, Y_2, \dots, Y_n$ are independent Poisson random variables with means $\lambda_1, \lambda_2, \dots, \lambda_n$.

How do you find the conditional probability function of $Y_1$ given that $\displaystyle\sum_{i=1}^n Y_i=m$?

I know that $\displaystyle\sum_{i=1}^n Y_i \sim Poisson(\displaystyle\sum_{i=1}^n \lambda_i)$.

I've never worked with conditional probabilities, so forgive me if the solution here is trivial...

user213065
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2 Answers2

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Ah this is quite nice since the answer that comes out is quite intuitive. Anyway you just apply Bayes' theorem, to get (for $0\leq k \leq m$):

$$ \begin{aligned} \Pr\left[Y_1 = k \bigg\vert \sum_{i=1}^n Y_i = m\right] &= \frac{\Pr\left[Y_1 = k \cap \sum_{i=1}^n Y_i = m\right]}{\Pr\left[\sum_{i=1}^n Y_i = m\right]} \\ &= \frac{\Pr\left[Y_1 = k \cap \sum_{i=2}^n Y_i = m-k\right]}{\Pr\left[\sum_{i=1}^n Y_i = m\right]} \\ &= \frac{\Pr\left[Y_1 = k \right]\Pr\left[\sum_{i=2}^n Y_i = m-k\right]}{\Pr\left[\sum_{i=1}^n Y_i = m\right]} \end{aligned} $$

In the final equality we used the independence of $Y_1$ and $\sum_{i=2}^k Y_i $.

Now you know all of the above expressions, since

$Y_1 \sim \mathrm{Pois}(\lambda_1)$, $\sum_{i=2}^n Y_i \sim \mathrm{Pois}(\sum_{i=2}^n \lambda_i)$ and $\sum_{i=1}^n Y_i \sim \mathrm{Pois}(\sum_{i=1}^n \lambda_i)$.

Plugging all of this in will finally yield:

$$ Y_1 \bigg\vert \sum_{i=1}^n Y_i = m \sim Bin\left(m, \frac{\lambda_1}{\sum_{i=1}^n \lambda_i}\right) $$

air
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    Technically, what you used was not Bayes' theorem, but the definition of conditional probability: $\Pr[A \mid B] = \Pr[A \cap B]/\Pr[B]$, whereas Bayes' is about reversing the conditionality: $$\Pr[A \mid B] = \Pr[B \mid A]\Pr[A]/\Pr[B].$$ – heropup Feb 04 '15 at 22:29
  • Woops, of course you are right. Thanks for pointing this out! – air Feb 04 '15 at 22:32
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You can assume that $Y_1 = n$ for whatever $n$ you want, and then compute the entire probability that the rest of the variables sum to $m - n$, by exploiting the fact that the sum of the rest of the variables is Poisson with mean given by the sum of means of the individual Poissons.

user2566092
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