# Thread: Quick Help with Joint Probability Mass function

1. ## Quick Help with Joint Probability Mass function

Trying to do this question.

Finding K i get 1/25 but I know this is wrong.

MY method P(1,1) + P(1,3) + P(2,3) = 1

2. Hello,

No, you're all correct !

Note : please next time, put your attachments as .jpg, not .bmp, as it's more painful to open !

3. Cool thanks! But I thought that P(1,1) = 1/3 and if K= 1/25

Then P(1,1) = 2/25?

Sorry about the pic format I usually change it but forgot this time.

4. Originally Posted by Niall101
Cool thanks! But I thought that P(1,1) = 1/3 and if K= 1/25

Then P(1,1) = 2/25?

Sorry about the pic format I usually change it but forgot this time.
Why would P(1,1)=1/3 ? Because there are only 3 points ?
This is only correct if we assume that the distribution is uniform.
Which is not necessarily the case.

Yes, P(1,1)=2/25

5. Cool Thanks so much! Im glad i was on the right track.

Does that same idea hold for when Im getting the marginal function of X

I get 2x^2 / 25 + 10/25 which is getting X over all values of Y where Y can be 1 or 3.

But is it also possible say for x = 1 and y = 50 for example which would still give a value for X even though the joint would give 0?

6. Perhaps it'll be clearer if you see it written this way :

Let $\displaystyle \mathcal{X}$ be the domain of X, namely {1,2}
Let $\displaystyle \mathcal{Y}$ be the domain of Y, namely {1,3}

Then for any $\displaystyle x\in\mathbb{R}$, we have :
$\displaystyle f_X(x)=P(X=x)=\sum_{y\in\mathcal{Y}} P(x,y)=P(x,1)+P(x,3)$

Which exactly gives what you got :
2x^2 / 25 + 10/25
if x is in $\displaystyle \mathcal{X}$

If $\displaystyle x\neq 1,2$, we still have $\displaystyle f_X(x)=P(x,1)+P(x,3)$, but then these values are 0, according to the definition of P(.,.)

7. Hey, thanks a lot for your help I appreciate it.