Statistical Averages question

[Gibson486]

Suppose i am given a function g(x) and f(x). g(x) is a function involving random variables. When I compute the Expected Value of Y, why do I use the values/equations of g(x)? Why can't I simply use f(x) as i did with the Expected value of X. I know doing this is the therom of Expected values of random value Y, but I do not see how it works and it still can be related to f(x).

 

[Heisenberg]

What are 'X' and 'Y'?

 

[DrPizza]

I don't quite understand your question... how are f(x) and g(x) related??
X? Y?

 

[TuxDave]

I'm glad I'm not the only one confused by the question.

 

[Random Variable]

I would be the completely wrong person to ask.

 

[Gibson486]

Sorry, I wrote the functions at the top wrong, but i figured it out. It was kind of obvious, but i just was kind of occupied at teh time i was lookign at it, so I really did not notice it.

Basically....

g(X) is the function of a random variable.

fX(x) is a function, maybe a PDF (probabilty density function)

The purpose is to find the expected value of y, but you are only given fX(x).

To find the expected value of x, you need to integrate x*fX(x) dx

To find expected value of y, you could integrate y*fY(y)dy or g(x)*fY(y) dy, however, you are not given fY(y).

However, there is a therom that states all you need to is integrate g(x)fX(x) dx.

I was wondering why we can use g(x)*f(x) dx to find expected value of y.

I found the answer while thinking about it in the car. Basically, g(x) is a function that maps Y to X. you could integrate y*fX(x) dx, but doing so would give you a y term, so to prevent that, we use what y is mapped/equal to.

 

[PandaBear]

Currently taking the same class (are you in EE grad school?)

For those who want to know more about this, fX(x) is the probability (pdf) of x, so if you want to know the expected value of x, you integrate its value times the probability at that value (hence x*fX(x)) over the entire domain.

Like you said, do the same for y, but you aren't given f(y) so you can't integrate y*fY(y). In your question g(X) is actually y=g(x), so you get to integrate g(x)*fY(y), now to get fY(y)..... let me check the book and answer later.

 

[Gibson486]

I am an undergrad...and I this class is busting my balls.....it sucks taking this and Linear Systems at the same time ( Ihad to do convolution without knowing what convolution was).

woah, I just read your response.......that what it's doing? dam...I need to pick your brain before i have my midterm on monday

 

[PandaBear]

I think I am taking the exact same classes as you (Linear system for some background I have to get, I came from another major).

In a nut shell, convolution is taking one function, flip it left side right, add a variable offset between it and another function, and add them along the whole domain (from -infinity to +infinity). Repeat the whole thing with the offset from -infinity to + infinity. Usually you are given a range to work with so you will not need to deal with infinite range. For undergrad you are probably only doing one side function (with u(t)) so when you flip one of the function, your range is like this:

-------------------------------|_______
*
_______|--------------------------------
=
_______|------------------|_________

Now, like everything else, the devil is in the detail.

 

[PandaBear]

Oh, about the fY(y), just found it.

Say your y = g(x) = ax + b and you know about Fy(Y), the cdf. Then x = (y-b) / a

dF/dy
= dF/dx * dx/dy (chain rule)
= dF/dx * d((y-b)/a) /dy (as mentioned above)
= dF/dx * d(y/a - b/a) / dy
= dF/dx * 1/a

Since pdf/pmf = dF/dy and dF/dx depends on which variable you are looking at.

fY(y) = 1/a * fX(x)


Of course, you have to do this all over again for a different y = g(x)