Expected Value in Joint Distribution

Is this correct?

E(XY) = (0x0x0.4) + (0x1x0.1) + (1x0x0.2) + (1x1x0.15) + (2x0x0.1) +(2x1x0.05)

Yep, you got it.

$E(XY)$ is simply $\sum xy\operatorname{P}(X=x,Y=y)$

For part d), you just need $\sum \operatorname{P}(X=x,Y=y)$, BUT you need to restrict your sum to the values of x,y which meet the given criterion in each case. For the second part you may find it easiest to work with the complementary event - or do it both ways for practise

Hi,

Thank you for your reply,

Am I correct in saying P(X>Y) = 0.5

and for the second part of D), do I ^3 the probabilities or the actual values?

I got 0.85 for the second part of question D, not sure if correct though

'fraid not. Which probabilities did you add, and why?



The values. You need to add the probabilities that correspond to the values of X,Y that meet the criterion.

I did 0.35 + 0.15 to get P(X>Y)

P(X=0) = 0.5
P(X=1) = 0.35
P(X=2) = 0.15

P(Y=0) = 0.7
P(Y=1) = 0.3

P(X=1) > P(Y=1) and P(X=2) > P(Y) so I added the probabilities together. Where did I go wrong?

Just had a thought and maybe its simply P(X=2) which is 0.15?

The last bit "P(X=1) > P(Y=1) and P(X=2) > P(Y)" is irrelevant, as you're interested in the values of X,Y, not their probabilities, when checking against the criterion.

You need to consider each cell individually.

E.g working along the top row.

First cell relates to X=0, Y=0. Is X>Y ? No. So, ignore it.
Next cell relates to X=1, Y=0. Is X>Y ? Yes, so we include the 0.2 in the sum of the probabilites.

And repeat for all 6 cells.

Ahhhh thank you now I get it . Cheers for all your help. I got 0.35 by the way