help with support of a function

• Feb 2nd 2011, 06:56 AM
tecne
help with support of a function
Hi,

I recently read your reply regarding the support of the function coming up from the addition of two random variables. I am referring to the following post:
sum of random variables - help to evaluate support of function

My problem is similar. I know that variable http://www.mathhelpforum.com/math-he...4e155c67a6.png is defined on http://www.mathhelpforum.com/math-he...97c2f26272.png and also that variable http://www.mathhelpforum.com/math-he...22904f345d.png is defined on http://www.mathhelpforum.com/math-he...7076be9f42.png. I am interested in the difference http://www.mathhelpforum.com/math-he...4acda1d7da.png. I am very confused however, since I think that the support of http://www.mathhelpforum.com/math-he...b808451dd7.png and hence the integration limits should be between http://www.mathhelpforum.com/math-he...bfbd2ae76a.png. On the other hand, I think that the appropriate limits might be http://www.mathhelpforum.com/math-he...afbd79ea51.png in the convolution integral. Therefore, my problem is determining the integration limits of the convolution integral in order to find http://www.mathhelpforum.com/math-he...3e467887dd.png.

Do you have any ideas? Your help will be greatly appreciated.

BR,

Alex
• Feb 2nd 2011, 11:34 AM
mr fantastic
Quote:

Originally Posted by tecne
Hi,

I recently read your reply regarding the support of the function coming up from the addition of two random variables. I am referring to the following post:
sum of random variables - help to evaluate support of function

My problem is similar. I know that variable http://www.mathhelpforum.com/math-he...4e155c67a6.png is defined on http://www.mathhelpforum.com/math-he...97c2f26272.png and also that variable http://www.mathhelpforum.com/math-he...22904f345d.png is defined on http://www.mathhelpforum.com/math-he...7076be9f42.png. I am interested in the difference http://www.mathhelpforum.com/math-he...4acda1d7da.png. I am very confused however, since I think that the support of http://www.mathhelpforum.com/math-he...b808451dd7.png and hence the integration limits should be between http://www.mathhelpforum.com/math-he...bfbd2ae76a.png. On the other hand, I think that the appropriate limits might be http://www.mathhelpforum.com/math-he...afbd79ea51.png in the convolution integral. Therefore, my problem is determining the integration limits of the convolution integral in order to find http://www.mathhelpforum.com/math-he...3e467887dd.png.

Do you have any ideas? Your help will be greatly appreciated.

BR,

Alex

If U = X - Y and http://www.mathhelpforum.com/math-he...97c2f26272.png and http://www.mathhelpforum.com/math-he...7076be9f42.png then the support of U will clearly be $\displaystyle \displaystyle -\infty < U < +\infty$ (in light of the extreme 'values' of X and Y, you should look at the extreme 'values' that it's possible for U to approach ....)
• Feb 2nd 2011, 02:19 PM
tecne
Thanks for your reply. In my case $\displaystyle U=Y-X$. However, I suspect that the same applies here and therefore $\displaystyle -\infty<U<\infty$. I do not know the extreme values in advance. Do you think I should plug in these limits directly to the convolution integral?

Thanks again.

BR,

Alex
• Feb 2nd 2011, 07:12 PM
mr fantastic
Quote:

Originally Posted by tecne
Thanks for your reply. In my case $\displaystyle U=Y-X$. However, I suspect that the same applies here and therefore $\displaystyle -\infty<U<\infty$. I do not know the extreme values in advance. Do you think I should plug in these limits directly to the convolution integral?

Thanks again.

BR,

Alex

• Feb 3rd 2011, 04:40 AM
Volga
I thought,

$\displaystyle u=y-x$, then $\displaystyle y=u+x$
$\displaystyle 0<y<\infty$ translate that for u:
$\displaystyle 0<u+x<\infty$
$\displaystyle -x<u<\infty$

So there should be at least one bound, line u=-x ???

Will wait for the full question...
• Feb 3rd 2011, 11:50 AM
tecne
Thanks for your replies. In my case $\displaystyle Z=Y-X$.

Let me begin with the basis of my problem, which is the cosine law. In more detail I have the following:

$\displaystyle r^2=y^2 + d^2 -2ydcos\gamma$.

The last term, i.e. $\displaystyle 2ydcos\gamma$ is $\displaystyle X$ and $\displaystyle y^2$ is $\displaystyle Y$. I know that $\displaystyle X$ is Gaussian distributed. Now, since $\displaystyle y$ denotes distance it can take any values from $\displaystyle [0,\infty)$. Also, $\displaystyle -1<cos\gamma<1$ and therefore I conclude that $\displaystyle -\infty<x<\infty$. On the other hand, $\displaystyle Y$ is Gamma distributed with $\displaystyle 0<y<\infty$.

Therefore, I am faced with the subtraction of two random variables, $\displaystyle Z= Y-X$. My aim of course is to find the density of $\displaystyle r^2$ but I approach the problem step by step. Having found $\displaystyle Z$, I then add the constant $\displaystyle d^2$, which will not affect the variance of the distribution but only its first order moment. I have made use of the following theorem:

$\displaystyle f_{Z}(z) = \int_{-\infty}^{\infty}f_{X,Y}(y-z,y)dy$. (1)

To proceed, I decompose the joint density function into the product of the conditional density of $\displaystyle f(X \vert Y)$ and the marginal $\displaystyle f(Y)$. My intuition suggests that the conditional density should be Gaussian. This is due to $\displaystyle X$ being Gaussian given $\displaystyle \sqrt{Y}$ being Nakagami. Of course, $\displaystyle Y$ will be Nakagami distributed if $\displaystyle Y^2$ is Gamma.

Now, obviously $\displaystyle 0<y<\infty$ (always positive), although this is not true for $\displaystyle X$ due to $\displaystyle cos\gamma$. Do you think that the support of $\displaystyle Z$ as applied to this case will still be $\displaystyle (-\infty,\infty)$, so that the integral in (1) is evaluated accordingly? I think that the integration limits should be between $\displaystyle [0,\infty)$ and that the support of $\displaystyle Z$ will be between $\displaystyle (-\infty,\infty)$. Please let me know your thoughts.

Thanks and apologies for the lengthy post.

BR,

Alex