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
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.
I was trying to simplify the integral,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?
Observe that the nature of the problem implies $\displaystyle 1 \le k<=T+N$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
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...
(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!
I don't know exactly what the details are, but its a detector of some kind.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
I must admit that I knew (or at least thought I knew) howOriginally 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
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.
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.