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

[s3a]

Homework Statement

The problem and its solution are attached in the TheProblemAndSolution.jpg file.

Homework 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

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

 

[SammyS]

Homework Statement

The problem and its solution are attached in the TheProblemAndSolution.jpg file.

I don't see that file anywhere .

Homework 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

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

 

[s3a]

Oops! :P

I just added it (to the opening post).

 

[Ray Vickson]

Homework Statement

The problem and its solution are attached in the TheProblemAndSolution.jpg file.

Homework 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

*********************************************************************************
Are you saying you are unable to do this integral? You ought to be able to do it almost without work; certainly you should not need a computer package. Once you have done the integral, you get another "easy" integral to do in order to find EX. Again, this is a simple integral of the type you must have done many times when you took introductory calculus.

BTW: since ## X,Y \geq 0##, all integrations run from ##0## to ##+\infty##, not from ##-\infty## to ##+\infty##. That is because the density vanishes for negative values of ##x## or ##y##.

**********************************************************

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

 

[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.

 

[Ray Vickson]

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.

OK, but what will you do on an exam?

 

[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. :)

 

[Ray Vickson]

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. :)

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.