This is a problem asked on the usenet group: sci.engr.radar+sonar

Show:

$\displaystyle

\int_0^{\infty} e^{-(T+N+1-k)y} (1-e^{-y})^{k-1} dy=

\frac{(k-1)!(T+N-k)!}{(T+N)!}

$

RonL

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- Mar 3rd 2006, 06:50 AMCaptainBlackFalse Alarm Integral
This is a problem asked on the usenet group: sci.engr.radar+sonar

Show:

$\displaystyle

\int_0^{\infty} e^{-(T+N+1-k)y} (1-e^{-y})^{k-1} dy=

\frac{(k-1)!(T+N-k)!}{(T+N)!}

$

RonL - Mar 3rd 2006, 09:07 AMThePerfectHacker
I wanted to call the term $\displaystyle T+N+1$ as $\displaystyle M$ but I am afraid to because maybe you mean that all three need to be positive integers in that case $\displaystyle M\geq 3$. Can you provide the domain for all the integers?

Maybe you can induct on $\displaystyle k$? That what I am trying now. - Mar 3rd 2006, 09:31 AMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**

$\displaystyle

\int e^{-(T+N+1-k)y} (1-e^{-y})^{k-1} dy

$

into an easier form. Notice it can be expressed as,

$\displaystyle

\int e^{-y}(1-e^{-y})^{k-1}(e^{-y})^{T+N-k}

$

Use the substitution, $\displaystyle u=1-e^{-y}$ then we have that $\displaystyle u'=e^{-y}$ and $\displaystyle (e^{-y})^{T+N-k}=(1-u)^{T+N-k}$. Thus, we have that,

$\displaystyle \int u'u^{k-1}(1-u)^{T+N-k}dx$ by the substitution rule we have that,

$\displaystyle \int u^{k-1}(1-u)^{T+N-k}du$. But I do not know if this form helps? - Mar 3rd 2006, 10:46 AMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

and $\displaystyle T+N>1$ amd these are all integers.

The first thing that I will do is simplify this a little by putting:

$\displaystyle M=T+N$

Then what we have to show becomes:

$\displaystyle

\int_0^{\infty} e^{-(M+1-k)y} (1-e^{-y})^{k-1} dy=

\frac{(k-1)!(M-k)!}{M!}

$

Let:

$\displaystyle

I_{k,M}=\int_0^{\infty} e^{-(M+1-k)y} (1-e^{-y})^{k-1} dy

$,

and proceed by integrating by parts:

$\displaystyle

I_{k,M}=\left[-\frac{(1-e^{-y})^{k-1}e^{-(M+1-k)y} }{M+1-k}\right]_0^\infty$$\displaystyle -\int_0^{\infty}\frac{e^{-(M+1-k)y}}{M+1-k}(k-1)e^{-y}(1-e^{-y})^{k-2}dy

$

Tiding this up:

$\displaystyle

I_{k,M}=\frac{k-1}{M+1-k}I_{k-1,M}

$

Then repeatedly using this recurrence and observing the end condition:

$\displaystyle

I_{1,M}=\frac{1}{M}

$

gives:

$\displaystyle

I_{k,M}=\frac{(k-1)!(M-k)!}{M!}

$

RonL - Mar 3rd 2006, 11:54 AMtopsquark
Hmmm...again I did it the hard way.

I expanded the integrand in powers of $\displaystyle e^{-y}$ and integrated. The integral is easy that way, but the form is nasty. However, I found a way to simplify using, of all things, Lagrange's Interpolation formula.

Just when you thought it was useless... :D

(Actually, to be honest I'm not quite there...I'm still off by a constant. But that ought to go away once I go through it again. I've got a method, that's what I was after.)

-Dan

BTW Thanks for the integral. I haven't had that much fun integrating for a while! - Mar 3rd 2006, 11:59 AMtopsquark
Since this came from an engineering usenet group, I presume this integral represents something. I'm curious to know if you had found that out.

-Dan - Mar 3rd 2006, 12:05 PMThePerfectHacker
I am curious has anyone tried to bring this to the Gamma Integral Form?

- Mar 3rd 2006, 12:05 PMCaptainBlackQuote:

Originally Posted by**topsquark**

I expect its something like an M from N detector, where a detection is flagged

if M threshold crossings are made out of N opportunities. Except of course this

has three parameters so presumably is more complicated than that.

Also somewhere in there the distribution of the noise is hidden (since this

is related to the calculation of the probability of false alarm)

RonL - Mar 3rd 2006, 12:14 PMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

to do this before I posted the question.

I posted it here as I wanted the solution on line and in mathematical

notation (it would be impossible to follow in plain ASCII), so I could post a

link to it in its original forum.

RonL - Mar 5th 2006, 06:10 PMJohn MullaneContext of the Integral
Hey Guys,

I posted that integral in the engineering forum, thanks alot for your help. Indeed it is a detector, for a radar, in my work, an FMCW radar. It is an equation to establish the detection threshold multiplier (T) to obtain a constant probablity of false alarm (Constant false alarm rate (CFAR) detector). In this case N, is the number of 'cells' used in the moving window function, and k is the kth cell, which is used to estimate the mean of the (Assumed exponentiallly distirubuted) noise. This appears in the Ordered-Statistics CFAR detector. The interested reader may look at :

ieeexplore.ieee.org/iel3/ 4347/12454/00573784.pdf?arnumber=573784

or, if you don't have access :

www.nato-asi.org/sensors2005/papers/rohling.pdf

I too believe it can be solved through the gamma function and its factorial properties.

Thanks again,

John. - Mar 5th 2006, 08:39 PMJohn MullaneFollow up question
If you are guys are interested, I'd also like to post the following integral which I'm also banging my head against. It is generally the second step in detector analysis - evaluating the probability of detection.

Show:

$\displaystyle

P_{D} = \int_{0}^{\infty}P[x\geq Tz|H_{1}]f_{Z}(z) dz = \bigg(1 + \displaystyle\frac{Tz}{1+S} \bigg)^{-N}

$

where

$\displaystyle

P[x\geq Tz|H_{1}] = \int_{Tz}^{\infty}f_{X}(x|H_{1})dx.

$

and,

$\displaystyle

f_{X}(x|H_{1}) = \displaystyle\frac{1}{\mu}e^{(\frac{-x}{\mu} +

S)}.I_{0}\bigg(2\sqrt{\displaystyle\frac{S x}{\mu}} \bigg)

$

where $\displaystyle I_{0} $ is a modified Bessel function of order zero. (This is a Ricean probability distribution function). S in this case, represents the average signal to noise ratio from the target.

As before , for an exponentially distributed Z, the probability density function of the kth element of N ordered (according to amplitude) independent samples of Z is given by:

$\displaystyle

f_{Z}(z) = \frac{k}{\mu}\binom{N}{k}(e^{\displaystyle{-z/\mu}})^{N-k+1}(1-e^{\displaystyle{-z/\mu}})^{k-1}.

$

As the cdf of a Ricean ($\displaystyle P[x\geq Tz|H_{1}] $) is generally evaluated in closed form using the Gaussian error (or complimentary error) function, I fail to see how $\displaystyle P_{D} $ can be obtained in the above form.

Any comments/ideas/suggestions are greatly appreciated.

Regards,

John. - Mar 5th 2006, 11:01 PMCaptainBlackQuote:

Originally Posted by**John Mullane**

RonL