# Poisson random variable question

• Mar 17th 2009, 09:46 PM
blues360
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.
• Mar 18th 2009, 08:18 AM
meymathis
So this is asking to find

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

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:
$P(\sum_{i=1}^{10} X_i = 20 | X_1 < 2) =P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 < 2) /P(X_1 < 2)$

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

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

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

Now everything is standard Poisson which you should be able to calculate.
• Mar 18th 2009, 05:57 PM
matheagle
switched the /\

Quote:

Originally Posted by meymathis
So this is asking to find

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

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:
$P(\sum_{i=1}^{10} X_i = 20 | X_1 < 2) =P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 < 2) /P(X_1 < 2)$

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

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

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

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

• Mar 22nd 2009, 07:29 PM
WaterMist
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!
• Mar 22nd 2009, 07:39 PM
matheagle
Your $\lambda$ seems to be 3.
So the distribution is $P(X=x)= e^{-\lambda}\lambda^x/x! = e^{-3}3^x/x!$
From that you can easily obtain $P(X=0)$ and $P(X=1)$
and $P(X<2)=P(X=0)+P(X=1)$.

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

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

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