Expected value of Rayleigh

  • Thread starter alman9898
  • Start date
1. The problem statement, all variables and given/known data
A density function used is:

[tex]f(y) = \frac{2y}{\theta}e^{\frac{-y^{2}}{\theta}} y > 0[/tex]

using method of transformations find:

(a) pdf of U = Y^2
(b) E(Y) and V(Y)

2. Relevant equations
f(u) = f(h^-1(u))*|dh^-1/du|
h^-1(u) = sqrt(u)

3. The attempt at a solution

The pdf Y^2 ~ exp(theta)

E(Y) = E(U^1/2).

That is:


So, multiply by sqrt(theta)/sqrt(theta) and group terms...


The integral is the Gamma function with parameter 3/2. So I get...

sqrt(PI/THETA)/2 as the answer....but a little internets research reveals the expected value *should* be sqrt(PI * THETA)/2. I can't find out where my math went wrong, it's something stupid but I've been stuck for a whole day...


Science Advisor
Homework Helper
Insights Author
Gold Member
Try changing the variable from u to s by letting [itex]s = u/\theta,\ du = \theta ds[/itex]:

[tex]\int_0^\infty\frac 1 \theta e^{-\frac u \theta}\, du=
\int_0^\infty\frac 1 \theta(\theta s)^{\frac 1 2}e^{-s}\theta\, ds =
\int_0^\infty\sqrt{\theta}s^{\frac 1 2}e^{-s}\,ds = \sqrt\theta\ \Gamma(\frac 3 2) [/tex]
That helps alot, but if it wasn't for checking my work I would've been surely wrong. Any idea why my method didn't work; it's a trick my professor uses to solve problems with 'known' solutions.


Science Advisor
Homework Helper
It's because the gamma function is the integral of sqrt(v)*exp(-v)*dv. Or the integral of sqrt(u/theta)*exp(-u/theta)*d(u/theta). Not the integral of sqrt(u/theta)*exp(-u/theta)*du. The last one is what you were putting to be gamma(3/2). That's not right. You are off by a factor of theta because you didn't change the integration variable. Probably happened because you were omitting the d(something) after your integrals.
Last edited:
Ok I understand it now. That clears up a lot, and explains the missing term I needed to make things all nice. (I would have another theta term in the numerator to cancel the one out in the denominator).

Want to reply to this thread?

"Expected value of Rayleigh" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads