Conditional Probability Distribution

[SP90]

Homework Statement

The Attempt at a Solution

I have that the joint probability mass function would be

$\Pi_{i=1}^{k} \frac{\lambda_{i}^{n_{i}}}{n_{i}!} e^{-\lambda_{i}}$

How would I go about applying the conditional to get the conditional distribution?

 

[Ray Vickson]

Homework Statement

The Attempt at a Solution

I have that the joint probability mass function would be

$\Pi_{i=1}^{k} \frac{\lambda_{i}^{n_{i}}}{n_{i}!} e^{-\lambda_{i}}$

How would I go about applying the conditional to get the conditional distribution?

You need to compute
$$P(N_1 = n_1, N_2 = n_2, \ldots, N_k = n_k | \sum_{i=1}^k N_i = n)<br /> = \frac{ P(N_1 = n_1, N_2 = n_2, \ldots, N_k = n_k \: \& \sum_{i=1}^k N_i = n)}{P(\sum_{i=1}^k N_i = n)}$$ for any k-tuple $(n_1, n_2, \ldots, n_k)$ that sums to n. Can you simplify the "event" in the numerator probability? Can you compute the denominator probability?

RGV

 

[SP90]

I don't understand how the $\sum_{i=1}^k N_i = n$ term constrains the joint distribution, given that each $\lambda_{i}$ can be distinct.

Also, to evaluate $P(\sum_{i=1}^k N_i = n)$, wouldn't this be the joint distribution, evaluated at $N_{1}=n-\sum_{i=2}^{k}N_{i}$ and then $N_{2}=n-\sum_{i=3}^{k}N_{i}$ and so on? I feel that's wrong because it doesn't seem like it'd collapse down nicely.

 

[Ray Vickson]

I don't understand how the $\sum_{i=1}^k N_i = n$ term constrains the joint distribution, given that each $\lambda_{i}$ can be distinct.

Also, to evaluate $P(\sum_{i=1}^k N_i = n)$, wouldn't this be the joint distribution, evaluated at $N_{1}=n-\sum_{i=2}^{k}N_{i}$ and then $N_{2}=n-\sum_{i=3}^{k}N_{i}$ and so on? I feel that's wrong because it doesn't seem like it'd collapse down nicely.

The $\lambda_i$ have absolutely nothing to do with the $n_i$; if you think they do, then you seriously misunderstand the material. The formula you wrote for the joint pmf is $P(N_1 = n_1, \ldots, N_k = n_k).$ The only role of the $\lambda_i$ is to control the size of this probability, but that has nothing to do with the $n_i$ values themselves.

RGV

 

[SP90]

But because the joint PMF is a multiplication of the individual PMFs, the term becomes $\lambda_{1}^{n_{1}}\lambda_{2}^{n_{2}}\lambda_{3}^{n_{3}}...$ which means the term $\sum_{i=1}^k N_i = n$ doesn't appear in the joint PMF (although it would if the $\lambda_{i}$ were equal), so I don't understand how the condition $\sum_{i=1}^k N_i = n$ is applied to give $P(N_1 = n_1, N_2 = n_2, \ldots, N_k = n_k \: \& \sum_{i=1}^k N_i = n)$

 

[Ray Vickson]

But because the joint PMF is a multiplication of the individual PMFs, the term becomes $\lambda_{1}^{n_{1}}\lambda_{2}^{n_{2}}\lambda_{3}^{n_{3}}...$ which means the term $\sum_{i=1}^k N_i = n$ doesn't appear in the joint PMF (although it would if the $\lambda_{i}$ were equal), so I don't understand how the condition $\sum_{i=1}^k N_i = n$ is applied to give $P(N_1 = n_1, N_2 = n_2, \ldots, N_k = n_k \: \& \sum_{i=1}^k N_i = n)$

Try first to figure out what happens when k = 2, then maybe k = 3. You should see a pattern emerging.

RGV