Rayleigh PDF Squared in the form of a Gamma PDF

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gman

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Hi All,
This is my first post. It looks like there are some pretty smart members here. I'm trying to find the PDF of the sum of the square of a set of N independant Rayleigh distributions each with the same variance. I've come across an interesting relationship at the bottom of the 'Rayleigh Distribution' wikipedia page stating:

\(\displaystyle \sum\limits_{i = 1}^N {R_i ^2 \left( {\sigma ^2 } \right)} = \Gamma \left( {N,2\sigma ^2 } \right)\)

I've tried to verify this for N=1, but am having some trouble...
For the left side of the equation I get (for N=1):

\(\displaystyle \begin{array}{l}
R\left( \sigma \right) = \frac{{xe^{ - x^2 /2\sigma ^2 } }}{{\sigma ^2 }} \\
R\left( {\sigma ^2 } \right) = \frac{{xe^{ - x^2 /2\sigma ^4 } }}{{\sigma ^4 }} \\
R^2 \left( {\sigma ^2 } \right) = \frac{{x^2 e^{ - x^2 /\sigma ^4 } }}{{\sigma ^8 }} \\
\end{array}\)

For the left side of the equation I get:

\(\displaystyle \begin{array}{l}
\Gamma \left( {x;k,\theta } \right) = x^{k - 1} \frac{{e^{ - x/\theta } }}{{\theta ^k \Gamma \left( k \right)}} \\
\Gamma \left( {x;k = N = 1,\theta = 2\sigma ^2 } \right) = \frac{{e^{ - x/2\sigma ^2 } }}{{2\sigma ^2 }} \\
\end{array}\)

The two sides of the equation (as I've calculated them anyway) aren’t equal. I figure I'm missing something...

\(\displaystyle R\left( {\sigma ^2 } \right) < > \Gamma \left( {x;k = N = 1,\theta = 2\sigma ^2 } \right)\)

Sorry about the <>, it's meant to signify the 'not equal' symbol

I believe wikipedia, but I don't know where I've gone wrong. Though the above shows the equations for N=1, my eventual goal is to calculate the cummulative distribution function of the sum of the square of N independant rayleigh PDFs of constant variance. It looks like it will be the form of the incomplete gamma function. Before attempting that I'd like to understand my mistake where N=1 though.

Thanks!
 
gman said:
Hi All,
This is my first post. It looks like there are some pretty smart members here. I'm trying to find the PDF of the sum of the square of a set of N independant Rayleigh distributions each with the same variance. I've come across an interesting relationship at the bottom of the 'Rayleigh Distribution' wikipedia page stating:

\(\displaystyle \sum\limits_{i = 1}^N {R_i ^2 \left( {\sigma ^2 } \right)} = \Gamma \left( {N,2\sigma ^2 } \right)\)

I've tried to verify this for N=1, but am having some trouble...
For the left side of the equation I get (for N=1):

\(\displaystyle \begin{array}{l}
R\left( \sigma \right) = \frac{{xe^{ - x^2 /2\sigma ^2 } }}{{\sigma ^2 }} \\
R\left( {\sigma ^2 } \right) = \frac{{xe^{ - x^2 /2\sigma ^4 } }}{{\sigma ^4 }} \\
R^2 \left( {\sigma ^2 } \right) = \frac{{x^2 e^{ - x^2 /\sigma ^4 } }}{{\sigma ^8 }} \\
\end{array}\)

For the left side of the equation I get:

\(\displaystyle \begin{array}{l}
\Gamma \left( {x;k,\theta } \right) = x^{k - 1} \frac{{e^{ - x/\theta } }}{{\theta ^k \Gamma \left( k \right)}} \\
\Gamma \left( {x;k = N = 1,\theta = 2\sigma ^2 } \right) = \frac{{e^{ - x/2\sigma ^2 } }}{{2\sigma ^2 }} \\
\end{array}\)

The two sides of the equation (as I've calculated them anyway) aren’t equal. I figure I'm missing something...

\(\displaystyle R\left( {\sigma ^2 } \right) < > \Gamma \left( {x;k = N = 1,\theta = 2\sigma ^2 } \right)\)

Sorry about the <>, it's meant to signify the 'not equal' symbol

I believe wikipedia, but I don't know where I've gone wrong. Though the above shows the equations for N=1, my eventual goal is to calculate the cummulative distribution function of the sum of the square of N independant rayleigh PDFs of constant variance. It looks like it will be the form of the incomplete gamma function. Before attempting that I'd like to understand my mistake where N=1 though.

Thanks!
Click to expand...
Hi, gman. The Wikipedia page says that \(\displaystyle \L R^2\) has a Gamma distribution, but you haven't calculated the distribution of \(\displaystyle \L R^2 .\) This is analogous to showing the square of a normal variable has a \(\displaystyle \L \chi^2\) distribution. It involves more work than you've shown.
 
The sum of the square of N independant standard normal PDFs is the chi squared distribution, and I'm able to find cumulative distribution function for that. What I'm stuck on is finding the CDF for the sum of N independant non-standard (mean = 0, but variance not equal 1) normal distributions.

Is there a relatively simple way to scale the CDF for the chi squared distribution to account for a normal distribution with non-unity variance?

Statistics is not my strong point, and any help with this problem is appreciated.

Thanks.
 
gman said:
The sum of the square of N independant standard normal PDFs is the chi squared distribution, and I'm able to find cumulative distribution function for that. What I'm stuck on is finding the CDF for the sum of N independant non-standard (mean = 0, but variance not equal 1) normal distributions.

Is there a relatively simple way to scale the CDF for the chi squared distribution to account for a normal distribution with non-unity variance?

Statistics is not my strong point, and any help with this problem is appreciated.

Thanks.
Click to expand...
Scale each squared normal variable by dividing by its variance. In symbols, if \(\displaystyle X_i \sim N(0,\sigma_i^2),\)
then \(\displaystyle X_i /\sigma_i \sim N(0,1)\) is a standard normal variable. So \(\displaystyle X_i^2/\sigma_i^2 = (X_i /\sigma_i)^2 \sim \chi^2(1)\) and \(\displaystyle \sum_{i=1}^N X_i^2 / \sigma_i^2 \sim \chi^2 (N) .\)
 
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