# Expected Value in Joint DistributionWatch

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Thread starter 5 years ago
#1
Hi all. Kind of struggling with C) and D)

I know how to calculate the marginal distribution, E(X), E(Y), VAR(X) but I don't know how to calculate E(XY). Can't seem to find anything online that helps either.

Do I have to calculate the probability that X+Y= 0, X+Y=1, X+Y=2 first and then find E(XY) through that?

For question d) I don't understand it and would appreciate if someone could clarify and give me some guidance on how I could go about answering it. How would I calculate the probability that random variable X is greater than Y?

Thanks 0
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Thread starter 5 years ago
#2
Is this correct?

E(XY) = (0x0x0.4) + (0x1x0.1) + (1x0x0.2) + (1x1x0.15) + (2x0x0.1) +(2x1x0.05)
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5 years ago
#3
(Original post by flabbyjoe)
Is this correct?

E(XY) = (0x0x0.4) + (0x1x0.1) + (1x0x0.2) + (1x1x0.15) + (2x0x0.1) +(2x1x0.05)
Yep, you got it. is simply For part d), you just need , 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 1
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Thread starter 5 years ago
#4
(Original post by ghostwalker)
Yep, you got it. is simply For part d), you just need , 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?
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Thread starter 5 years ago
#5
I got 0.85 for the second part of question D, not sure if correct though 0
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5 years ago
#6
(Original post by flabbyjoe)
Hi,

Thank you for your reply,

Am I correct in saying P(X>Y) = 0.5
'fraid not. Which probabilities did you add, and why?

and for the second part of D), do I ^3 the probabilities or the actual values?
The values. You need to add the probabilities that correspond to the values of X,Y that meet the criterion.
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5 years ago
#7
(Original post by flabbyjoe)
I got 0.85 for the second part of question D, not sure if correct though Yep. Agree with that.
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Thread starter 5 years ago
#8
(Original post by ghostwalker)
'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?
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Thread starter 5 years ago
#9
Just had a thought and maybe its simply P(X=2) which is 0.15?
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5 years ago
#10
(Original post by flabbyjoe)
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?
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.
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5 years ago
#11
(Original post by flabbyjoe)
Just had a thought and maybe its simply P(X=2) which is 0.15?
See my previous post.
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Thread starter 5 years ago
#12
(Original post by ghostwalker)
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 0
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5 years ago
#13
(Original post by flabbyjoe)
Ahhhh thank you now I get it . Cheers for all your help. I got 0.35 by the way Yep, I agree with that. 0
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5 years ago
#14
Sorry OP but the title of this thread didn't leave me thinking about maths. Ignore me.

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