Exponential Density distribution

Hi Guy's,
In the Exponential Density distribution

cxe^(-x^2/2)

I have been asked to find the Expected value of x and determine the value of c


I think that the Expected value is 2.

But I'm not sure how to calculate the value of c.

If I take the rate parameter (lambda) of this part of the function (-x^2/2) to = 1/2, Do I need c to be a value that makes sure that cx is always = 1/2?

regards
Brendan

I'm not really sure what you're asking, a number of ideas seem to be a bit muddled together. I could piece most of it together, but I have no real idea where you're going with the rate parameter business, since this isn't an exponential distribution. The condition you've proposed on c wouldn't seem to make sense, since it would imply c = 1/(2x), that is, c is variable.

What's the probability density function for the random variable you're working with? You said the "exponential density distribution" is cxe^(-x^2/2), perhaps you mean that's the PDF?

How did you get the expected value you said, particularly if you didn't know how to find c? [Using the function given as a PDF I get Sqrt(pi/2) instead of 2, if you want to check your work with someone else's]

As for determining c, what properties must a PDF possess? Using a pointed example, is the function x^2 a proper PDF--why isn't it? Similarly, is the constant function 3.48+0x a proper PDF, and if not why?

Thanks for your reply,

I have been given this question.

The Random variable X follows the distribution given by the density
f(x) = cxe^(-x^2/2)

determine c and E(x)

In my lecture notes there is formula that say mu = alpha x beta
As alpha = 1 and beta = 2 which is the devisor in ^(-x^2/2)

I think I may need to read up a bit more on this

Thanks again Brendan

I'm not sure which distribution your alpha and beta are coming from, but it's likely one that doesn't match this PDF. It's probably best to read up on this more, and if needed ask your teacher for help since there seem to be a few conceptual holes that are best plugged face-to-face.

If you're interested, this appears to be the Rayleigh distribution. I hesitated to post the link but if you can figure out the answers to the question you posted from the info on that page, you should (probably ;P) have a good enough conceptual understanding anyway. If you use that page alone, just be sure you understand why the sigma is in the denominator of the PDF.

Thanks,
To find the value of C.

Do you take the integral of cxe^(-x^2/2) between 0 and infinity and solve this to = 1 and if so would this make C = 1, or do you take the integral between 0 and 1 ?

Brendan

You take the integral over all valid values of your random variable X. In this case it's almost certainly 0 to infinity. At the least, stopping at 1 would be very arbitrary--you may as well stop at 2, or e, or pi, or a zillion.

I was wondering are you getting 1 as the value of C?

Yes.

Using the Rayleigh distribution I linked, you can also see sigma (squared) = 1 = 1 / C -> C = 1.

Thanks Mate!