# Thread: Conditional Expectation question... No idea what to do

1. ## Conditional Expectation question... No idea what to do

Let Y  Bin(4; p) and assume that the conditional distribution of X given that Y = y is
Poi(y). Write down E(X/Y = y). Use the formula E(X) = E(E(X/Y )) to nd E(X).

Have no idea where tostart...

2. Originally Posted by sebjory
Let Y Bin(4; p) and assume that the conditional distribution of X given that Y = y is
Poi(y). Write down E(X/Y = y). Use the formula E(X) = E(E(X/Y )) to nd E(X).

Have no idea where tostart...
Start with the formula: $E(X | Y = y) = \sum_x x \cdot f(x | y)$.

3. thanks very much for your help.

how ever, i did get to this bit but i found it tough to integrat ethe poisson distribution in this way... i feel im missing something ...

4. Originally Posted by sebjory
thanks very much for your help.

how ever, i did get to this bit but i found it tough to integrat ethe poisson distribution in this way... i feel im missing something ...
It's a sum, not an integral. Please show the work you've done and where you get stuck.

5. of course the poisson distribution is discrete...

however this is exactly where im having trouble- what exactly does this summation mean and how do use it?

6. Originally Posted by sebjory
of course the poisson distribution is discrete...

however this is exactly where im having trouble- what exactly does this summation mean and how do use it?
You're meant to recognise that when you substitute the conditional pdf into $E(X | Y = y) = \sum_x x \cdot f(x | y)$ the summation is the definition of the mean of a Poisson random variable ....

7. which is 'y' in this case...

therefore E(X) is 4p?

8. Originally Posted by sebjory
which is 'y' in this case...

therefore E(X) is 4p?
It would appear so.