[SOLVED] Density function

• Oct 3rd 2008, 04:41 AM
approx
[SOLVED] Density function
I've got this problem:
f(y) = $\displaystyle (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha}$, y>0
0, elsewhere

Now I'm supposed to find $\displaystyle E(Y^k)$ for any positive integer k.

I start out like this:

$\displaystyle E(Y^k)$ = $\displaystyle \int_0^\infty y^{k}f(y) dy$ = $\displaystyle \int_0^\infty y^{k} (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha} dy$

I then move out $\displaystyle (1/\alpha)$ and m outside the integral =

$\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k} y^{m-1}e^{{-y}^{m}/\alpha} dy = \displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy$

and from here I don't know how to go on. The correct answer is $\displaystyle \Gamma (k/m+1)\alpha ^{k/m}$

and I know that $\displaystyle \Gamma (\alpha) = \int_0^\infty y^{\alpha-1}e^{-y} dy$

but I don't see how $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy can be rewritten to the correct answer. Thanks in advance for your help. • Oct 3rd 2008, 02:26 PM mr fantastic Quote: Originally Posted by approx I've got this problem: f(y) = \displaystyle (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha}, y>0 0, elsewhere Now I'm supposed to find \displaystyle E(Y^k) for any positive integer k. I start out like this: \displaystyle E(Y^k) = \displaystyle \int_0^\infty y^{k}f(y) dy = \displaystyle \int_0^\infty y^{k} (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha} dy I then move out \displaystyle (1/\alpha) and m outside the integral = \displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k} y^{m-1}e^{{-y}^{m}/\alpha} dy$ = $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy and from here I don't know how to go on. The correct answer is \displaystyle \Gamma (k/m+1)\alpha ^{k/m} and I know that \displaystyle \Gamma (\alpha) = \int_0^\infty y^{\alpha-1}e^{-y} dy but I don't see how \displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy$ can be rewritten to the correct answer. Thanks in advance for your help.

• Oct 5th 2008, 05:03 AM
approx
mr fantastic: I'm sorry to say that I still don't understand how to rewrite that expression into the right answer.
• Oct 5th 2008, 05:11 AM
mr fantastic
Quote:

Originally Posted by approx
mr fantastic: I'm sorry to say that I still don't understand how to rewrite that expression into the right answer.

The first three lines of the reference I've given you are crystal clear I would have thought. What exactly don't you understand?
• Oct 5th 2008, 05:40 AM
approx
Thanks for your fast answer. I don't understand which substitutions I'm supposed to do. Should I let:

$\displaystyle u = y^m/\alpha$ ?

which gives $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-u}} du ? • Oct 5th 2008, 05:54 PM mr fantastic Quote: Originally Posted by approx Thanks for your fast answer. I don't understand which substitutions I'm supposed to do. Should I let: \displaystyle u = y^m/\alpha ? which gives \displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-u}} du$ ?

Your substitution is incorrect for a number of reasons:

1. You still have y in the integral, there should only be u's.
2. You have not substituted the correct exprssion for dy. $\displaystyle dy \neq du$ !

Note that $\displaystyle u = \frac{y^m}{\alpha} \Rightarrow dy = \frac{\alpha}{m \, y^{m-1}}\, du$ and $\displaystyle y = (\alpha \, u)^{1/m}$.

It's expected that at this level you can correctly substitute a change of variable in an integral.
• Oct 6th 2008, 06:08 AM
approx
Ok. So I've done some thinking and came up with this substitution:

$\displaystyle u = y^m$, which gives $\displaystyle y = u^{1/m}$ and $\displaystyle dy = (1/m) u^{{(1/m)}-1} du$. Right?

And then I go on:

$\displaystyle (m/\alpha) \int_0^\infty u^{({1/m})^{k+m-1}}e^{-u/\alpha}(1/m) u^{(1/m)-1}du$

I move $\displaystyle 1/m$ outside which gives $\displaystyle (1/m)(m/\alpha) = (1/\alpha)$ outside the integral.

$\displaystyle (1/\alpha) \int_0^\infty u^{({1/m})^{k+m-1}}e^{-u/\alpha}(1/m) u^{(1/m)-1}du$

=

$\displaystyle (1/\alpha) \int_0^\infty u^{(({(k-1)}/m)+1)}e^{-u/\alpha} u^{(1/m)-1}du$

I put together the u:s and receives:

$\displaystyle (1/\alpha) \int_0^\infty u^{k/m}e^{-u/\alpha} du$

Am I right so far?
• Oct 6th 2008, 11:59 AM
mr fantastic
Quote:

Originally Posted by approx
Ok. So I've done some thinking and came up with this substitution:

$\displaystyle u = y^m$, which gives $\displaystyle y = u^{1/m}$ and $\displaystyle dy = (1/m) u^{{(1/m)}-1} du$. Right?

And then I go on:

$\displaystyle (m/\alpha) \int_0^\infty u^{({1/m})^{k+m-1}}e^{-u/\alpha}(1/m) u^{(1/m)-1}du$

I move $\displaystyle 1/m$ outside which gives $\displaystyle (1/m)(m/\alpha) = (1/\alpha)$ outside the integral.

$\displaystyle (1/\alpha) \int_0^\infty u^{({1/m})^{k+m-1}}e^{-u/\alpha}(1/m) u^{(1/m)-1}du$

=

$\displaystyle (1/\alpha) \int_0^\infty u^{(({(k-1)}/m)+1)}e^{-u/\alpha} u^{(1/m)-1}du$

I put together the u:s and receives:

$\displaystyle (1/\alpha) \int_0^\infty u^{k/m}e^{-u/\alpha} du$

Am I right so far?

Yes. Now substitute $\displaystyle w = \frac{u}{\alpha} \Rightarrow u = \alpha w$.
• Oct 7th 2008, 05:45 AM
approx
Thank you! I got the right answer after the last substitution.