# Thread: Poisson random variable question

1. ## Poisson random variable question

Let X and Y be independent Poisson random variables with means L and u respectively. Then it is known that X+Y is a Poisson with means L+u. Now let X1,...,X10 be independent Poisson random variables with the same mean, L=3. Find the probability that X1+...+X10=20 when you already know that X1<2.

2. So this is asking to find

$\displaystyle P(\sum_{i=1}^{10} X_i = 20 | X_1 < 2)$

So break it down. We try to work this stuff out into independent events so that we can calculate the pieces.

Standard equality for conditional probability:
$\displaystyle P(\sum_{i=1}^{10} X_i = 20 | X_1 < 2) =P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 < 2) /P(X_1 < 2)$

Break numerator into disjoint sets.
$\displaystyle = (P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 =0)+P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 =1))/P(X_1 < 2)$

Now that $\displaystyle X_1$ is fixed in the two summands, you can write the expressions as what happens to $\displaystyle X_2,\ldots,X_{10}$ and what happens to $\displaystyle X_1$ separately.
$\displaystyle = (P(\sum_{i=2}^{10} X_i = 20 \textrm{ and } X_1 =0)+P(\sum_{i=2}^{10} X_i = 19 \textrm{ and } X_1 =1))/P(X_1 < 2)$

Now we have "and"'s between independent events.
$\displaystyle = (P(\sum_{i=2}^{10} X_i = 20)P(X_1 =0)+P(\sum_{i=2}^{10} X_i = 19)P(X_1 =1))/P(X_1 < 2)$

Now everything is standard Poisson which you should be able to calculate.

3. switched the /\

Originally Posted by meymathis
So this is asking to find

$\displaystyle P(\sum_{i=1}^{10} X_i = 20 | X_1 < 2)$

So break it down. We try to work this stuff out into independent events so that we can calculate the pieces.

Standard equality for conditional probability:
$\displaystyle P(\sum_{i=1}^{10} X_i = 20 | X_1 < 2) =P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 < 2) /P(X_1 < 2)$

Break numerator into disjoint sets.
$\displaystyle = (P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 =0)+P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 =1))/P(X_1 < 2)$

Now that $\displaystyle X_1$ is fixed in the two summands, you can write the expressions as what happens to $\displaystyle X_2,\ldots,X_{10}$ and what happens to $\displaystyle X_1$ separately.
$\displaystyle = (P(\sum_{i=2}^{10} X_i = 20 \textrm{ and } X_1 =0)+P(\sum_{i=2}^{10} X_i = 19 \textrm{ and } X_1 =1))/P(X_1 < 2)$

Now we have "and"'s between independent events.
$\displaystyle = (P(\sum_{i=2}^{10} X_i = 20)P(X_1 =0)+P(\sum_{i=2}^{10} X_i = 19)P(X_1 =1))/P(X_1 < 2)$

Now everything is standard Poisson which you should be able to calculate.

4. Could you please guide me to the solution?

I'm having trouble doing the "standard poisson" part, and am getting a result of 1.

Thank you!

5. Your $\displaystyle \lambda$ seems to be 3.
So the distribution is $\displaystyle P(X=x)= e^{-\lambda}\lambda^x/x! = e^{-3}3^x/x!$
From that you can easily obtain $\displaystyle P(X=0)$ and $\displaystyle P(X=1)$
and $\displaystyle P(X<2)=P(X=0)+P(X=1)$.

Now for the rv $\displaystyle S=\sum_{i=2}^{10}X_i$ well thats a sum of nine Poisson's so its distribution is...

$\displaystyle P(S=s)= e^{-9\lambda}(9\lambda)^s/s! = e^{-27}(27)^s/s!$

Plug in 19 and 20 for s, and that's all folks.