Ah ok, thank you! I'm on to part
(c) Find E(X), E(Y ), E(Y |X) and V ar(X|Y ).
I get the following for E(X) which I think is correct:
But when I do it for E(y) I run into a problem on the 6th step:
I don't know what to do with the X in a(1/2 - x1/2 - 0 - 0). I need to solve for a single number for E(y) but the x is giving me problems.
Hmm. How do I have the X value already out? Im not really following.
The joint density isn't uniform. The marginal of Y is uniform. The joint depends on the value of x so it can't be uniform.
This can't be a marginal distribution for Y. Why not? Well it's a function of x... and the marginal distribution needs to be a function of y only.
Edit: Also, you already know what 'a' is. Stop referring to 'a' and instead replace any reference of 'a' with 2.
So what would the marginal distribution for Y be then? I think BGM confused me more.
lol sorry to confuse you.
Actually I think that in the process of solving the normalizing constant ,
you should already have a rough idea about the marginal distributions of
and respectively.
Anyway, let see if you agree with the following steps. You are almost got
everything correct (except some places do not substitute back, I do not know why).
Just reply if you cannot accept/agree with any one of them.
1. The solution
2. Relationship between the joint distribution and marginal distribution:
3. The given joint distribution:
4. Combining 2 and 3 gives
5. Integration of polynomials:
6. Integration is linear:
7. Combining 5 and 6 gives
8. So the conclusion is
and
If you are familiar with the definition of independence,
then you should know are independent if and only if
provided that their joint and marginal density exists.
So by observations, if the joint density can be written as
for some functions
independent of and independent of ,
(of course the support is somehow "rectangular", independent of ,
e.g. in this case)
then we know that
and are independent.
Thats the reason I say that by observation, without really doing integration,
you should be confident to say they are independent. Also, since the joint
distribution is independent from the dummy variable of ,
the marginal pdf of is a constant function (on [0, 1]),
so the marginal distribution of must be a uniform distribution.
That all makes sense, thank you. So back to my question regarding E(Y)
(c) Find E(X), E(Y ), E(Y |X) and V ar(X|Y ).
I get the following for E(X) which I think is correct:
But when I do it for E(y) I run into a problem on the 6th step:
I don't know what to do with the X in a(1/2 - x1/2 - 0 - 0). I need to solve for a single number for E(y) but the x is giving me problems.
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