Thread: probability and expected value

A have a question whre the random variables x and y have the following joint probability density:



I resolve the constant a = 6/5, so ,

What the expected value of 1 / X, given that Y = 3/2.?

tks

Your notation is confusing. What is "se"? You should not use the same letters for random variables and function argument.

Hi Mathman,

Sorry, "se" = "for".

tks

I get a = 1/2. Could you show me your derivation?

We can say that??

So,









a=6/5 (?)

Also check your boundary conditions, the original density function was from 0 to 2 but you are integrating from 1 to 2?

Dear river_rat and mathman,

Sorry, the original question is from 1 to 2 and not 0 to 2 as:

Ok, given that I get ?

Hi river_rat

My development gave my 6/5 (see post of 11 nov). Something wrong in my reasoning?

The bounds on the inner integral look wrong, remember that Y >= X

a=3/2

right?

That looks good

Have you calculated the conditional expectation yet?

Not yet... today. Tks!!

18/10?

Now have 2 other topics to resolve:

(b)Define the marginal probability density function py (Y).
answer: 18/10 (ok?)

(c) Determine o valor esperado de 1/X, dado que Y = 3/2.
answer: ??

Do a sanity check on your answer:

which implies that and so no matter how we assign positive weight to that interval we cannot get an average less that or greater than 1. So we know there is something wrong with your answer it would seem.