# How is E(X) = 1 obtained for this continuous joint pdf?

1. Mar 13, 2015

### s3a

1. The problem statement, all variables and given/known data
The problem and its solution are attached in the TheProblemAndSolution.jpg file.

2. Relevant equations
E(X) = integral from -infinity to infinity of x * $f_X(x)$

$f_X(x)$ = integral from -infinity to infinity of $f_{XY}(x,y)$ dy

3. The attempt at a solution
Basically, how is E(X) = 1 (and E(Y) = 1) obtained?

I tried the following, but I don't get E(X) = 1.:
1. http://www.wolframalpha.com/input/?i=integrate+e^(-x-y)+dy+from+0+to+x

2. http://www.wolframalpha.com/input/?i=cosh(x)-cosh(2+x)-sinh(x)+sinh(2+x)&lk=1&a=ClashPrefs_*Math-

3. http://www.wolframalpha.com/input/?i=integrate+x+e^(-x)+dx+from+0+to+inf

4. http://www.wolframalpha.com/input/?i=integrate+x+*+(+-e^(-2+x)+e^(-x)+)+dx+from+0+to+infinity

#### Attached Files:

• ###### TheProblemAndSolution.jpg
File size:
12 KB
Views:
72
2. Mar 13, 2015

### SammyS

Staff Emeritus
I don't see that file anywhere .

3. Mar 13, 2015

### s3a

Oops! :P

I just added it (to the opening post).

4. Mar 13, 2015

### Ray Vickson

5. Mar 18, 2015

### s3a

Actually, I was very tired on that day, so I made stupid mistakes, and I think I get it now.

As for the limits of integration, I'd say the interval (in its non-simplified form) is from negative infinity to positive infinity if you use the notation $f_{XY}(x,y)$ instead of $e^{-x-y}$, because $f_{XY}(x,y)$ is 0 when it is not the case that x > 0 and y > 0, but I think we're saying the same thing in different words.

Having said that, thanks for your input. :)

P.S.
I used Wolfram Alpha, not because I'm not capable of computing integrals, but because my priority was to deal with the problem-solving nature of the problem and not with the algebraic component of the problem.

6. Mar 18, 2015

### Ray Vickson

OK, but what will you do on an exam?

7. Mar 22, 2015

### s3a

I do end up computing the integrals, but I do so after having figured out the problem-solving aspect of the problem. I find that that backwards approach works better for me.

Thanks for caring, though. :)

8. Mar 22, 2015

### Ray Vickson

The solution (from the book) that you posted makes heavy weather out of something that is actually very simple. Your bivariate f(x,y) is a product of two univariate functions (one of x and the other of y) so the random variable are automatically independent. Thus, covariance = 0, etc---end of story.

I guess the point I am trying to get across is that 0 correlation is a general property of independent random variables, and does not depend in any way on the detailed forms of the densities. Somewhere (perhaps in your textbook or perhaps in your course notes) that fact would, I hope, have been established, and if that is so then doing it all over again in this special case is like re-inventing the wheel. If that has not been presented to you yet, then I guess you do need to write down the integrals; however, you would not need to actually do the integrals to get an explicit final answer. All you need is to establish is $E(X Y) = EX \, EY$, without actually needing to compute $EX$ and $EY$ explicitly.

Of course, on an exam you need to do whatever the examiner tells you to do, but it is still a good habit to take provable, correct shortcuts wherever you can do so.