# Conditional Expectation of Poisson Random Variables

Suppose $X_1,X_2,\ldots,X_n$ is a random sample from a Poisson Distribution with mean $\theta$. How can I find the conditional expectation $E \left( X_1+X_2+3X_3 |\sum_{i=1}^n X_i \right)$?

I know that $\sum X_i$ has a $poisson (n\theta$) distribution. Similarly the random variable $X_1+X_2+X_3$ has a $poisson (3\theta)$ distribution. I get confused with the required summations afterwards though.

Thank you.

• Due to the linearity of the expected value, and the i.i.d (=exchangeability) assumption, the expected value you want is equivalent to $5E(X_i\mid \sum_{i=1}^nX_i)$ for any $i$. – Alecos Papadopoulos Jan 18 '14 at 20:09
• @AlecosPapadopoulos Thanks but what is the joint density of these two random variables? – JohnK Jan 18 '14 at 20:27
• The conditional distribution itself is a known result. Look up the Wikipedia article on Poisson, and let me know of any remaining issues. – Alecos Papadopoulos Jan 18 '14 at 20:31

Define $S_n=\sum_{i=1}^n X_i$. By symmetry, $$\mathrm{E}\left[ X_1 \mid S_n \right] = \mathrm{E}\left[ X_2 \mid S_n \right] = \dots = \mathrm{E}\left[ X_n \mid S_n \right] \quad \textrm{a.s.} \quad (*)$$ Hence, using $(*)$ and the linearity of the conditional expectation, we have $$\mathrm{E}\left[ X_1 \mid S_n \right] = \frac{1}{n} \mathrm{E}\left[ X_1+\dots+X_n \mid S_n \right] = \frac{1}{n} \mathrm{E}\left[ S_n \mid S_n \right] = \frac{S_n}{n} \quad \textrm{a.s.}$$ The same reasoning leads to $$\mathrm{E}\left[ X_1 +X_2 +3X_3\mid S_n \right] = 5\,\mathrm{E}\left[ X_1 \mid S_n \right] = \frac{5\,S_n}{n} \quad \textrm{a.s.}$$ Now, remember that $S_n\sim \mathrm{Poisson}(n\theta)$, and find the pmf of $5\,S_n/n$ (consider its support).

• Thank you. Can you please explain further what you have done in step 2? I do not understand why $E [X_1|S_n ] =S_n /n$ – JohnK Jan 19 '14 at 0:06
• Can you use $(*)$ to prove that $\mathrm{E}\left[ X_1 \mid S_n \right] = \frac{1}{n} \mathrm{E}\left[ X_1+\dots+X_n \mid S_n \right]$? – Zen Jan 19 '14 at 0:09
• Yep. It makes sense, thank you. I was asking about the $E[S_n|S_n]=S_n$. That is a cool identity. – JohnK Jan 19 '14 at 0:11
• That's one of the properties of the conditional expectation. Informally, if you have the information $S_n$, then your best guess for $S_n$ is exactly $S_n$. – Zen Jan 19 '14 at 0:14
• The important thing here is to understand that $\mathrm{E}[X]$ is a real number, while $\mathrm{E}[X\mid Y]$ is a random variable. Take alook at this question: stats.stackexchange.com/questions/38700/… – Zen Jan 19 '14 at 0:15

The OP has apparently found the way, so I am posting an answer.

I will denote $Z \equiv \sum_{i=1}^n X_i$. By linearity of the expected value we have $$E \left( X_1+X_2+3X_3 |Z \right)= E \left( X_1 |Z \right)+E \left( X_2 |Z \right)+3E \left(X_3 |Z \right)$$

Since the variables are i.i.d. they are also exchangeable,at least with respect to $Z$ (to which they have a symmetric relationship), so the three conditional expected values will be equal:

$$E \left( X_1+X_2+3X_3 |Z \right)= 5E \left( X_1 |Z \right)$$

Moreover, it is a known result that the conditional distribution of $X_1$ conditional on $Z=k$ is a Binomial,

$$X_1 | Z=k \sim Bin\left(k, \frac {E(X_1)}{E(Z)}\right) = Bin\left(k, 1/n\right)$$

and so

$$5E \left( X_1 |Z=k \right) = 5\frac kn$$

Our conditional expectation is viewed as a function of $Z$, is not conditioned just on $Z$ acquiring a specific value. Generalizing the last equation we obtain

$$5E \left( X_1 |Z \right) = \frac 5n Z= 5 \frac 1n \sum_{i=1}^n X_i$$

Note that

$$E \left( X_1 |Z \right) \rightarrow_p E(X_1) \;\;\text {as}\;\; n\rightarrow \infty$$

which should be intuitive.

• You might consider revising slightly the sentence beginning with "Since the variables are i.i.d....". The fact that they are i.i.d. does not imply that they are exchangeable conditional on an arbitrary $Z$. This $Z$ is special! – cardinal Jan 19 '14 at 1:38
• @cardinal I thought it self-understood but I will point it out, sure. Thanks. – Alecos Papadopoulos Jan 19 '14 at 1:43
• Thank you very much. It's helpful to have two ways to arrive at the result. – JohnK Jan 19 '14 at 10:04