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sanctifier

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## Homework Statement

If random variables [itex] X [/itex] and [itex] Y [/itex] are independent and both belong to Possion distribution of parameters [itex] \lambda_1 [/itex] and [itex] \lambda_2 [/itex], then what is the conditional distribution of [itex] X [/itex] when the condition [itex] X + Y = m[/itex] is given?

## Homework Equations

Possion distribution of parameter [itex] \lambda [/itex]: [itex] P(x)= \frac{\lambda^x}{x!} e^{-\lambda} [/itex]

## The Attempt at a Solution

Answer:

Because [itex] P(X=x|X+Y=m) = \frac{P(X=x \cup X+Y=m)}{P(X+Y=m)} [/itex]

Let [itex] \begin{cases}u=x \\v=x+y \end{cases} [/itex] then [itex] \begin{cases}x=u \\y=v-u \end{cases} [/itex]

[itex] Jacobian= \begin{bmatrix} \frac{dx}{du} & \frac{dx}{dv} \\ \frac{dy}{du} & \frac{dy}{dv} \end{bmatrix} = \begin{bmatrix}1 & 0 \\-1 & 1 \end{bmatrix} = 1 [/itex]

The joint distribution of [itex] X [/itex] and [itex] Y [/itex] is [itex] g(x,y)= \frac{\lambda_1^x}{x!} \frac{\lambda_2^y}{y!}e^{-\lambda_1 - \lambda_2} [/itex]

Then [itex]f(u,v)=g(u,v-u)|Jacobian|=\frac{\lambda_1^u}{u!} \frac{\lambda_2^{v-u}}{(v-u)!}e^{-\lambda_1 - \lambda_2} [/itex]

Hence the marginal distribution of [itex] v [/itex] is:

[itex] f_v(v)=\sum_{u=0}^{m}f(u,v)=e^{-\lambda_1-\lambda_2}\lambda_2^v \sum_{u=0}^{m} {( \frac{\lambda_1}{\lambda_2} )}^u \frac{1}{u!(v-u)!} [/itex]

[itex] P(X=x|X+Y=m)=P(U=u|V=m)= \frac{f(u,m)}{f_v(m)} [/itex]

Does the sum [itex] \sum_{u=0}^{m} {( \frac{\lambda_1}{\lambda_2} )}^u \frac{1}{u!(v-u)!} [/itex] have a concise form?

Is the answer correct? Thank you in advance!