Probability: What is the conditional distribution of X?

[sanctifier]

Homework Statement

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

Homework Equations

Possion distribution of parameter $\lambda$: $P(x)= \frac{\lambda^x}{x!} e^{-\lambda}$

The Attempt at a Solution

Answer:

Because $P(X=x|X+Y=m) = \frac{P(X=x \cup X+Y=m)}{P(X+Y=m)}$

Let $\begin{cases}u=x \\v=x+y \end{cases}$ then $\begin{cases}x=u \\y=v-u \end{cases}$

$Jacobian= \begin{bmatrix} \frac{dx}{du} & \frac{dx}{dv} \\ \frac{dy}{du} & \frac{dy}{dv} \end{bmatrix} = \begin{bmatrix}1 & 0 \\-1 & 1 \end{bmatrix} = 1$

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

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

Hence the marginal distribution of $v$ is:

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

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

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

Is the answer correct? Thank you in advance!

 

[kduna]

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

Is the answer correct? Thank you in advance!

Notice that:$\frac{1}{u!(v-u)!} = \frac{1}{v!} {v \choose u}$.

When finding the marginal, you should be summing to infinity, which actually just means summing through v (since v choose u will be zero thereafter).

Therefore your sum is: $\frac{1}{v!} \sum_{u=0}^{v} {v \choose u} {(\frac{\lambda_1}{\lambda_2})}^u 1^{v-u} = \frac{1}{v!}(\frac{\lambda_1}{\lambda_2} + 1)^v$.

Here I used the binomial formula. We can further simplify this though.

$\frac{1}{v!}(\frac{\lambda_1}{\lambda_2} + 1)^v = \frac{1}{v!} (\frac{\lambda_1 + \lambda_2}{\lambda_2})^v$.

Putting this into your formula for the marginal of $X+Y$ we see that $X+Y$ is Poisson with parameter $\lambda_1 + \lambda_2$. In fact this holds for any finite sum of Poisson random variables, and you just proved the base case for induction.

 

[sanctifier]

This is correct.

kduna, thank you a lot!

 

[Ray Vickson]

Homework Statement

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

Homework Equations

Possion distribution of parameter $\lambda$: $P(x)= \frac{\lambda^x}{x!} e^{-\lambda}$

The Attempt at a Solution

Answer:

Because $P(X=x|X+Y=m) = \frac{P(X=x \cup X+Y=m)}{P(X+Y=m)}$

Let $\begin{cases}u=x \\v=x+y \end{cases}$ then $\begin{cases}x=u \\y=v-u \end{cases}$

$Jacobian= \begin{bmatrix} \frac{dx}{du} & \frac{dx}{dv} \\ \frac{dy}{du} & \frac{dy}{dv} \end{bmatrix} = \begin{bmatrix}1 & 0 \\-1 & 1 \end{bmatrix} = 1$

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

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

Hence the marginal distribution of $v$ is:

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

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

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

Is the answer correct? Thank you in advance!

You made some errors.
(1)
$$P(X=x|X+Y=m) = \frac{P(X=x \cup X+Y=m)}{P(X+Y=m)} \longleftarrow \text{ wrong}$$
Can you spot the mistake?

(2) Jacobians and all those things are wrong for this problem; they apply to continuous random variables having probability density functions, but in your case all the random variables are discrete and do not have densities at all---instead, they have probability mass functions.

 

[sanctifier]

You made some errors.
(1)
$$P(X=x|X+Y=m) = \frac{P(X=x \cup X+Y=m)}{P(X+Y=m)} \longleftarrow \text{ wrong}$$
Can you spot the mistake?

(2) Jacobians and all those things are wrong for this problem; they apply to continuous random variables having probability density functions, but in your case all the random variables are discrete and do not have densities at all---instead, they have probability mass functions.

To (1)

Did you mean $$P(X=x|X+Y=m) = \frac{P(X=x \bigcap X+Y=m)}{P(X+Y=m)}$$

To (2)

Then what shall I do if the standard means doesn't work?

 

[Ray Vickson]

To (1)

Did you mean $$P(X=x|X+Y=m) = \frac{P(X=x \bigcap X+Y=m)}{P(X+Y=m)}$$

To (2)

Then what shall I do if the standard means doesn't work?

Standard means DO work; just use them correctly. For integer m, can you compute $P(X+Y=m)?$ For integer $x, m$ can you compute $P(X = x \; \& \; X+Y=m)?$

Hint: how would you express the event $\{ X+Y = m\}$ in terms of events like $\{ X = j \}$ and $\{ Y = k \}$? You came very close to having it in your original post, but you did not finish the job.

It was your use of Jacobians, etc., that was wrong, not some of the final formulas you got. In other words, you almost had the right answers for the wrong reasons. Doing that on an exam would still cost you points.

 

[sanctifier]

Thank you for your insistent reply.

As you suggested,

$P(X=x \cap X+Y=m)=P(X=x \cap Y=m-x)= \frac{\lambda_1^x}{x!}e^{-\lambda_1} *\frac{\lambda_2^{m-x}}{(m-x)!} e^{-\lambda_2}=\lambda_2^m e^{-\lambda_1-\lambda_2} (\frac{\lambda_1}{\lambda_2})^x \frac{1}{m!} {m \choose x}$

$P(X+Y=m) = \sum_{x=0}^{m} \frac{\lambda_1^x}{x!} e^{-\lambda_1} \frac{\lambda_2^{m-x}}{(m-x)!} e^{-\lambda_2} =\lambda_2^m e^{-\lambda_1-\lambda_2}\sum_{x=0}^m (\frac{\lambda_1}{\lambda_2} )^x \frac{1}{m!} {m \choose x} = \lambda_2^m e^{-\lambda_1-\lambda_2} \frac{1}{m!} ( \frac{\lambda_1}{\lambda_2}+1 )^m$

$P(X=x|X+Y=m) = \frac{P(X=x \cap X+Y=m)}{P(X+Y=m)} = \frac{P(X=x \cap Y=m-x)}{P(X+Y=m)} = \frac{(\frac{\lambda_1}{\lambda_2})^x {m \choose x} }{ ( \frac{\lambda_1}{\lambda_2}+1 )^m}$

Is this correct?

 

[Ray Vickson]

Thank you for your insistent reply.

As you suggested,

$P(X=x \cap X+Y=m)=P(X=x \cap Y=m-x)= \frac{\lambda_1^x}{x!}e^{-\lambda_1} *\frac{\lambda_2^{m-x}}{(m-x)!} e^{-\lambda_2}=\lambda_2^m e^{-\lambda_1-\lambda_2} (\frac{\lambda_1}{\lambda_2})^x \frac{1}{m!} {m \choose x}$

$P(X+Y=m) = \sum_{x=0}^{m} \frac{\lambda_1^x}{x!} e^{-\lambda_1} \frac{\lambda_2^{m-x}}{(m-x)!} e^{-\lambda_2} =\lambda_2^m e^{-\lambda_1-\lambda_2}\sum_{x=0}^m (\frac{\lambda_1}{\lambda_2} )^x \frac{1}{m!} {m \choose x} = \lambda_2^m e^{-\lambda_1-\lambda_2} \frac{1}{m!} ( \frac{\lambda_1}{\lambda_2}+1 )^m$

$P(X=x|X+Y=m) = \frac{P(X=x \cap X+Y=m)}{P(X+Y=m)} = \frac{P(X=x \cap Y=m-x)}{P(X+Y=m)} = \frac{(\frac{\lambda_1}{\lambda_2})^x {m \choose x} }{ ( \frac{\lambda_1}{\lambda_2}+1 )^m}$

Is this correct?

Yes, it is correct. Good work. However, it can all be made neater and more revealing.

First, you should note that your formula for $P(X+Y=m)$ simplifies to
$$P(X+Y=m) = \frac{\mu^m e^{-\mu}}{m!}, \;\; \mu = \lambda_1 + \lambda_2,$$
so $X+Y = \text{Poisson}(\lambda_1 + \lambda_2)$. This is a standard result.

Next: notice that your final formula for $P(X=k|X+Y=m)$ can be written as
$$P(X=k | X+Y=m) = {m \choose k} p^k (1-p)^{m-k}, \;\; p = \frac{\lambda_1}{\lambda_1+\lambda_2},$$
so $X | X+Y = m$ is binomial, with parameters $m$ and $p = \lambda_1/(\lambda_1+\lambda_2)$.

 

[sanctifier]

Thank you very much, Ray.