How to Solve a Probability Problem with Independent Stochastic Variables?

[Hummingbird25]

Hi People,

I have this Probability Problem which is given me a headach,

Given two independent Stochastic variables (X, Y) where X is Poisson distributed $$Po(\lambda)$$ and Y is Poisson distributed $$Po(\mu)$$.

Where $$\lambda$$, $$\mu > 0$$. Let $$m \geq 0$$ and $$p = \frac{\lambda}{\lambda+ \mu}$$

By the above I need to show, that

$$P(X=l| X+Y=m) = \left( \begin{array}{cc} l \\ m \end{array} \right ) p^l (1-p)^{(l-m)}$$

Proof:

Its know that

$$\left( \begin{array}{cc} l \\ m \end{array} \right ) = \frac{l!}{m!(1-m)!}$$

Wherefore the Binormal formula can be written as

$$\frac{l!}{(m!(l-m))!} p^l (1-p)^{(l-m)}$$

for $$m \geq 0$$ I get:$$\frac{-(\frac{\lambda + \mu}{\mu})^{l} \cdot p^{l} \cdot \mu}{l!(l!-1) \cdot (\lambda + \mu)}$$

Any surgestions on how to processed from here?

Sincerely Yours
Hummingbird25

 

[HallsofIvy]

No, no suggestions on how to proceed from there since you seem to be going at it backwards.

For any probability distribution, P(A and B)= P(A|B)*P(B) so that
P(A|B)= P(A and B)/P(B).

You are looking for P(X= l| X+ Y= m) so A is "X= l" and B is "X+ Y= m".
Then A and B is "X= l and X+ Y= m" which is the same as "X= l and Y= m-l". Now P(A and B)= P(X= l)*P(Y= m) .

 

[Hummingbird25]

Hi Hall,

Hello since X and Y is Poisson distributed then

$$P(x=l) * P(Y=m) = \frac{\lambda^l}{l !} e^{-\lambda} * \frac{\lambda^m}{m !} e^{-\mu}$$

Is this the next?

Sincerley Yours
Hummingbird25

 

[pizzasky]

I think HallsofIvy made a small mistake in his last statement. He probably meant "P(A and B)= P(X= l)*P(Y= m-l)".

 

[Hummingbird25]

Hello since X and Y is Poisson distributed then

$$P(X=l) * P(Y=m-l) = \frac{\lambda^l}{l !} e^{-\lambda} *$$ (what do I need to add here then)?

Is this the next?

Sincerley Yours
Hummingbird25

I think HallsofIvy made a small mistake in his last statement. He probably meant "P(A and B)= P(X= l)*P(Y= m-l)".

 

[pizzasky]

$$P(X=l) * P(Y=m-l) = \frac{\lambda^l}{l !} e^{-\lambda} * \frac{\mu^{m-l}}{(m-l)!} e^{-\mu}$$

And Hummingbird25, can I request that you check the question again? I tried solving the question and my answer was $$P(X=l| X+Y=m) = \left( \begin{array}{cc} m \\ l \end{array} \right ) p^l (1-p)^{(m-l)}$$. Perhaps there is something wrong with my working...

 

[Hummingbird25]

Hello Pizzasky and thank You,

I looked at my assigment again I and discovered, that You result was the right one, I had (by mistake swapped l and m.

The right answer as You formulated:

$$\left( \begin{array}{cc} m \\ l \end{array} \right ) p^l (1-p)^{(m-l)}$$Sincerely Hummingbird25

$$P(X=l) * P(Y=m-l) = \frac{\lambda^l}{l !} e^{-\lambda} * \frac{\mu^{m-l}}{(m-l)!} e^{-\mu}$$

And Hummingbird25, can I request that you check the question again? I tried solving the question and my answer was $$P(X=l| X+Y=m) = \left( \begin{array}{cc} m \\ l \end{array} \right ) p^l (1-p)^{(m-l)}$$. Perhaps there is something wrong with my working...

 

[Hummingbird25]

One last thing Pizzasky,

Which formula did You use to achive the binormal formula?

Sincerely Yours
Hummingbird25

$$P(X=l) * P(Y=m-l) = \frac{\lambda^l}{l !} e^{-\lambda} * \frac{\mu^{m-l}}{(m-l)!} e^{-\mu}$$

And Hummingbird25, can I request that you check the question again? I tried solving the question and my answer was $$P(X=l| X+Y=m) = \left( \begin{array}{cc} m \\ l \end{array} \right ) p^l (1-p)^{(m-l)}$$. Perhaps there is something wrong with my working...

 

[pizzasky]

Reply

You can use the substitution $$p = \frac{\lambda}{\lambda+ \mu}$$ to get the Binomial formula. Also, try to figure out what expression "1-p" corresponds to.