How Do You Compute the MLE for p in a Treatment Effectiveness Study?

[ianrice]

Homework Statement

1. An experiment consists of giving a sequences of patients a risky treatment until two have died, and then recording N, the number who survived. If p is the proportion killed by the treatment, then the distribution of N is:
P(N=n)=(n+1)(1-p)ⁿ p²

1)Find a general formula for the MLE for p:
2)the experiment is done in 8 hospitals and the observed values of N are 3,0,4,2,3,5,1,3. Compute the estimate for the p derived in part 1

Homework Equations

2. I know I have to start by finding a general form of ∏(n_i+1)(1-p)n_i p² with n_i where i goes from 1 to m.

The Attempt at a Solution

3. So I get hung up on how to handle the (n_i+1) term.

 

[Ray Vickson]

Homework Statement

1. An experiment consists of giving a sequences of patients a risky treatment until two have died, and then recording N, the number who survived. If p is the proportion killed by the treatment, then the distribution of N is:
P(N=n)=(n+1)(1-p)ⁿ p²

1)Find a general formula for the MLE for p:
2)the experiment is done in 8 hospitals and the observed values of N are 3,0,4,2,3,5,1,3. Compute the estimate for the p derived in part 1

Homework Equations

2. I know I have to start by finding a general form of ∏(n_i+1)(1-p)n_i p² with n_i where i goes from 1 to m.

The Attempt at a Solution

3. So I get hung up on how to handle the (n_i+1) term.

^{What is preventing you from computing P(N1=3,N2=0,N3=4,...,N8=3)?}

 

[ianrice]

What is preventing you from computing P(N1=3,N2=0,N3=4,...,N8=3)?

The question asks for the general form of the MLE of p. then to compute P(N1,N2...N8). I need help with the general form

 

[D H]

Homework Equations

2. I know I have to start by finding a general form of ∏(n_i+1)(1-p)^{n_i p2 with n_i where i goes from 1 to m.}

There's a typo in your post. Witness that hanging .

I suspect you meant $\Pi_{i=1}^m (n_i+1)(1-p)^{n_i} p^2$

The Attempt at a Solution

3. So I get hung up on how to handle the (n_i+1) term.

This isn't that hard. Let's start with m=1, i.e., one observation. Then that product is simply $(n_1+1)(1-p)^{n_1}p^2$. With m=2, the product becomes $(n_1+1)(1-p)^{n_1}p^2\,(n_2+1)(1-p)^{n_2}p^2$. Try collecting terms. You should get something of the form $A (1-p)^B p^C$, where A, B, and C are related to the observed values. Now generalize to m observations.

Where does this reach a maximum?

 

[Ray Vickson]

The question asks for the general form of the MLE of p. then to compute P(N1,N2...N8). I need help with the general form

Start with simple cases, by writing the formula for the likelihood function of P(n1,n2) and maybe P(n1,n2,n3). This will show you the pattern that applies as well to larger problems.

 

[ianrice]

There's a typo in your post. Witness that hanging <sup>.

I suspect you meant $\Pi_{i=1}^m (n_i+1)(1-p)^{n_i} p^2$
This isn't that hard. Let's start with m=1, i.e., one observation. Then that product is simply $(n_1+1)(1-p)^{n_1}p^2$. With m=2, the product becomes $(n_1+1)(1-p)^{n_1}p^2\,(n_2+1)(1-p)^{n_2}p^2$. Try collecting terms. You should get something of the form $A (1-p)^B p^C$, where A, B, and C are related to the observed values. Now generalize to m observations.

Where does this reach a maximum?</sup>

<sup>So I've collected terms for m=3 and get the following:

(n₁+1)(n₂+1)(n₃+1)(1-p)n₁(1-p)n₂(1-p)n₃p2p2p2

So by using your A(1-p)BpC:

I think B=∑n_i
C=2n
A is where I'm still stuck. Am I missing something completely I can't seem to figure out how to reduced (n₁+1)(n₂+1)(n₃+1)

And to find the maximum I need to set the derivative to 0 and find p the parameter the question ask for.</sup>

 

[D H]

What are $(1-p)^{n_1}(1-p)^{n_2}(1-p)^{n_3}$ and $p^2p^2p^2$? Surely you can reduce these to a simpler form!

 

[ianrice]

What are $(1-p)^{n_1}(1-p)^{n_2}(1-p)^{n_3}$ and $p^2p^2p^2$? Surely you can reduce these to a simpler form!

I got those:

$(1-p)^{n_1}(1-p)^{n_2}(1-p)^{n_3}$ = (1-p)^∑n_i

$p^2p^2p^2$ = p²ⁿ

 

[Ray Vickson]

So I've collected terms for m=3 and get the following:

(n₁+1)(n₂+1)(n₃+1)(1-p)^n₁(1-p)^n₂(1-p)^n₃p²p²p²

So by using your A(1-p)^Bp^C:

I think B=∑n_i
C=2n
A is where I'm still stuck. Am I missing something completely I can't seem to figure out how to reduced (n₁+1)(n₂+1)(n₃+1)

And to find the maximum I need to set the derivative to 0 and find p the parameter the question ask for.

Why would you care about reducing $(n_1+1) (n_2+1) \cdots (n_m+1)$? It has no effect whatsoever on the optimal value of p. In other words, the p that maximizes f(p) is the same p that maximizes 1000*f(p) or (1/100)*f(p) or any K*f(p) for constant K.

 

[ianrice]

Why would you care about reducing $(n_1+1) (n_2+1) \cdots (n_m+1)$? It has no effect whatsoever on the optimal value of p. In other words, the p that maximizes f(p) is the same p that maximizes 1000*f(p) or (1/100)*f(p) or any K*f(p) for constant K.

Ok right I understand. I was confused.

So am I correct with the following next steps:


L(p)= ∏(n_i+1)(1-p)^∑n_ip²ⁿ

Then in the next take is to find l(p) which is l(p)=Ln(L(p))

So I end up with:

l(p)= Ln(∏(n_i+1))+∑n_iLn(1-p)+2n*Ln(p)


Then taking the derivative of dl(p)/dp and setting that to zer
dl(p)/dp= 0+ \frac{∑n}{1-p} +\frac{2n}{p}

 

[Ray Vickson]

Ok right I understand. I was confused.

So am I correct with the following next steps:


L(p)= ∏(n_i+1)(1-p)^∑n_ip²ⁿ

Then in the next take is to find l(p) which is l(p)=Ln(L(p))

So I end up with:

l(p)= Ln(∏(n_i+1))+∑n_iLn(1-p)+2n*Ln(p)


Then taking the derivative of dl(p)/dp and setting that to zer
dl(p)/dp= 0+ \frac{∑n}{1-p} +\frac{2n}{p}

Almost OK, but $d \ln(1-p) /dp = -1/(1-p)$, not $+1/(1-p)$. The thing you wrote could never = 0 for any p between 0 and 1.