Expert: Paul Klarreich - 5/11/2006

Question
Hi I have been pondering this a while, and I am sure its incredibly obvious, but maybe you can show me the light at the end of the tunnel!

I am maximising a basic likelihood function:

L(N|n,p)=(N)p^n(1-p)^N-n
        (p)
I have converted to the log likelihood to maximise, and taken derivative assuming the approximation:


cln(N!)/dN=ln(N)

which gives
dl/dN=ln(N)-ln(N-n)+ln(1-p)

I cannot for the life of me solve for N, which is apparently

N=n/p

help!

Will

Answer
Hi, Bill,
(or is it Will?)
 
You wrote:
Subject:  likelihood function
Question:  Hi I have been pondering this a while, and I am sure its incredibly obvious, but maybe you can show me the light at the end of the tunnel!

I am maximising a basic likelihood function:

L(N|n,p)=(N)p^n(1-p)^N-n
      (p)
I have converted to the log likelihood to maximise, and taken derivative assuming the approximation:

cln(N!)/dN=ln(N)

which gives
dl/dN=ln(N)-ln(N-n)+ln(1-p)

I cannot for the life of me solve for N, which is apparently N=n/p

help!

Will
-----------------------------
I don't know what you are talking about, exactly, but snce you said you were MAXIMIZING something, and then found its derivative, the next step would be to set that derivative equal to zero and solve.

Set  ln(N)-ln(N-n)+ln(1-p) = 0

Now use some basic logarithm properties;  
ln A - ln B = ln(A/B),  and  
ln A + ln B = ln(AB), etc. to write:

  N(1 - p)
ln -------- = 0
   N - n

And one more property:  ln X = 0  means  X = 1

  N(1 - p)
 -------- = 1
   N - n

N(1 - p) = N - n
N - Np = N - n
 - NP = -n
   Np = n
    N = n/p
Does that do it?