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Math Help - False Alarm Integral

  1. #1
    Grand Panjandrum
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    False Alarm Integral

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

    Show:

    <br />
\int_0^{\infty} e^{-(T+N+1-k)y} (1-e^{-y})^{k-1} dy=<br />
\frac{(k-1)!(T+N-k)!}{(T+N)!}<br />

    RonL
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  2. #2
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    I wanted to call the term T+N+1 as M but I am afraid to because maybe you mean that all three need to be positive integers in that case M\geq 3. Can you provide the domain for all the integers?

    Maybe you can induct on k? That what I am trying now.
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  3. #3
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    Quote Originally Posted by CaptainBlack
    This is a problem asked on the usenet group: sci.engr.radar+sonar

    Show:

    <br />
\int_0^{\infty} e^{-(T+N+1-k)y} (1-e^{-y})^{k-1} dy=<br />
\frac{(k-1)!(T+N-k)!}{(T+N)!}<br />

    RonL
    I was trying to simplify the integral,
    <br />
\int e^{-(T+N+1-k)y} (1-e^{-y})^{k-1} dy<br />
    into an easier form. Notice it can be expressed as,
    <br />
\int e^{-y}(1-e^{-y})^{k-1}(e^{-y})^{T+N-k}<br />
    Use the substitution, u=1-e^{-y} then we have that u'=e^{-y} and (e^{-y})^{T+N-k}=(1-u)^{T+N-k}. Thus, we have that,
    \int u'u^{k-1}(1-u)^{T+N-k}dx by the substitution rule we have that,
    \int u^{k-1}(1-u)^{T+N-k}du. But I do not know if this form helps?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack
    This is a problem asked on the usenet group: sci.engr.radar+sonar

    Show:

    <br />
\int_0^{\infty} e^{-(T+N+1-k)y} (1-e^{-y})^{k-1} dy=<br />
\frac{(k-1)!(T+N-k)!}{(T+N)!}<br />

    RonL
    Observe that the nature of the problem implies 1 \le k<=T+N
    and T+N>1 amd these are all integers.


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

    M=T+N

    Then what we have to show becomes:

    <br />
\int_0^{\infty} e^{-(M+1-k)y} (1-e^{-y})^{k-1} dy=<br />
\frac{(k-1)!(M-k)!}{M!}<br />

    Let:

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

    and proceed by integrating by parts:

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

    Tiding this up:

    <br />
I_{k,M}=\frac{k-1}{M+1-k}I_{k-1,M}<br />

    Then repeatedly using this recurrence and observing the end condition:

    <br />
I_{1,M}=\frac{1}{M}<br />

    gives:

    <br />
I_{k,M}=\frac{(k-1)!(M-k)!}{M!}<br />

    RonL
    Last edited by CaptainBlack; March 3rd 2006 at 10:51 AM.
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  5. #5
    Forum Admin topsquark's Avatar
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    Hmmm...again I did it the hard way.

    I expanded the integrand in powers of 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!
    Last edited by topsquark; March 3rd 2006 at 11:55 AM. Reason: Addendum
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  6. #6
    Forum Admin topsquark's Avatar
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    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
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    I am curious has anyone tried to bring this to the Gamma Integral Form?
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by topsquark
    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
    I don't know exactly what the details are, but its a detector of some kind.
    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
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack
    This is a problem asked on the usenet group: sci.engr.radar+sonar

    Show:

    <br />
\int_0^{\infty} e^{-(T+N+1-k)y} (1-e^{-y})^{k-1} dy=<br />
\frac{(k-1)!(T+N-k)!}{(T+N)!}<br />

    RonL
    I must admit that I knew (or at least thought I knew) how
    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
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    Context 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.
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    Follow 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:
    <br />
P_{D} = \int_{0}^{\infty}P[x\geq Tz|H_{1}]f_{Z}(z) dz  = \bigg(1 + \displaystyle\frac{Tz}{1+S} \bigg)^{-N}<br />
    where
    <br />
P[x\geq Tz|H_{1}] = \int_{Tz}^{\infty}f_{X}(x|H_{1})dx.<br />

    and,
    <br />
f_{X}(x|H_{1}) = \displaystyle\frac{1}{\mu}e^{(\frac{-x}{\mu} +<br />
S)}.I_{0}\bigg(2\sqrt{\displaystyle\frac{S x}{\mu}} \bigg)<br />
    where  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:
    <br />
f_{Z}(z) = \frac{k}{\mu}\binom{N}{k}(e^{\displaystyle{-z/\mu}})^{N-k+1}(1-e^{\displaystyle{-z/\mu}})^{k-1}.<br />

    As the cdf of a Ricean (  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  P_{D}  can be obtained in the above form.

    Any comments/ideas/suggestions are greatly appreciated.
    Regards,
    John.
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  12. #12
    Grand Panjandrum
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    Quote Originally Posted by John Mullane
    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.
    Thanks for the reference, it looks interesting.

    RonL
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