Joint distribution function & conditional excpected value

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Thomas74

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Good afternoon, I'm working on my homework, but since I'm not really sure about what I have done, I want somebody to check it for me.
Thank you!

Random vector X=(x1,x2) is created by two independent random variables x1 and x2. Knowing probability distribution functions of these variables, find out: Joint distribution function of x, the expected value (x1,x2), and conditional expected values of (x1|x2).

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a) What do the sum of probabilities of the distribution must equal? You can use this as a check to see if a) is correct
b)The question asking for E(X1∗X2)E(X_1*X_2) not E(X1,X2)E(X_1,X_2). The second latter expression doesn't make sense, so no comma. You did set up the solution correctly. Also, since the random variables are independent, then E[X1X2]=E[X1]E[X2]E[X_1X_2]=E[X_1]E[X_2]. Try to find E[X1]E[X2]E[X_1]E[X_2] and see if you get the same answer as a check.
c) Looks fine
 
Last edited:
BigBeachBananas said:
a) What do the sum of probabilities of the distribution must equal? You can use this as a check to see if a) is correct
b)The question asking for E(X1∗X2)E(X_1*X_2) not E(X1,X2)E(X_1,X_2). The second latter expression doesn't make sense, so no comma. You did set up the solution correctly. Also, since the random variables are independent, then E[X1X2]=E[X1]E[X2]E[X_1X_2]=E[X_1]E[X_2]. Try to find E[X1]E[X2]E[X_1]E[X_2] and see if you get the same answer as a check.
c) Looks fine
Click to expand...
a) yes, I think it has to be equal
b) I got the same result by multiplying E[X1]E[X2]=0.8X(−0.2)=−0.16E[X_1]E[X_2]= 0.8 X (-0.2)= -0.16
Thank you guys @BigBeachBananas and @blamocur
 
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