Mixing
Parameter estimation using method of maximum likelihood. What am I doing wrong?
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Question from Michael Baron's Probability and Statistics for Computer Scientists, 2nd edition:
Data: 3 7 5 3 2. Assume data is produced from Geometric distribution. Estimate $p$, the geometric distribution parameter. (Geometric distribution: $P(x) = (1-p)^{x-1}p$, $x=1,2,...$)
My solution, using method of maximum likelihood: Find $M = sum_{i=1}^5 ln(P(x_i))$. Then solve for $frac{dM}{dp}=0$.
- $M = sum_{i=1}^5 ln((1-p)^{x_i - 1}p)=15ln(1-p)+ 5ln p$
- $frac{dM}{dp}=0=frac{15}{1-p}+frac{5}{p} to p=-frac{1}{2}$
This is obviously wrong, as $p$ is a probability. I did this twice, what am I doing wrong?
maximum-likelihood
asked Jan 8 at 9:39
an apprenticean apprentice
978
$endgroup$
add a comment |
$begingroup$
Question from Michael Baron's Probability and Statistics for Computer Scientists, 2nd edition:
Data: 3 7 5 3 2. Assume data is produced from Geometric distribution. Estimate $p$, the geometric distribution parameter. (Geometric distribution: $P(x) = (1-p)^{x-1}p$, $x=1,2,...$)
My solution, using method of maximum likelihood: Find $M = sum_{i=1}^5 ln(P(x_i))$. Then solve for $frac{dM}{dp}=0$.
- $M = sum_{i=1}^5 ln((1-p)^{x_i - 1}p)=15ln(1-p)+ 5ln p$
- $frac{dM}{dp}=0=frac{15}{1-p}+frac{5}{p} to p=-frac{1}{2}$
This is obviously wrong, as $p$ is a probability. I did this twice, what am I doing wrong?
maximum-likelihood
asked Jan 8 at 9:39
an apprenticean apprentice
978
$endgroup$
add a comment |
$begingroup$
Question from Michael Baron's Probability and Statistics for Computer Scientists, 2nd edition:
Data: 3 7 5 3 2. Assume data is produced from Geometric distribution. Estimate $p$, the geometric distribution parameter. (Geometric distribution: $P(x) = (1-p)^{x-1}p$, $x=1,2,...$)
My solution, using method of maximum likelihood: Find $M = sum_{i=1}^5 ln(P(x_i))$. Then solve for $frac{dM}{dp}=0$.
- $M = sum_{i=1}^5 ln((1-p)^{x_i - 1}p)=15ln(1-p)+ 5ln p$
- $frac{dM}{dp}=0=frac{15}{1-p}+frac{5}{p} to p=-frac{1}{2}$
This is obviously wrong, as $p$ is a probability. I did this twice, what am I doing wrong?
maximum-likelihood
asked Jan 8 at 9:39
an apprenticean apprentice
978
$endgroup$
Question from Michael Baron's Probability and Statistics for Computer Scientists, 2nd edition:
Data: 3 7 5 3 2. Assume data is produced from Geometric distribution. Estimate $p$, the geometric distribution parameter. (Geometric distribution: $P(x) = (1-p)^{x-1}p$, $x=1,2,...$)
My solution, using method of maximum likelihood: Find $M = sum_{i=1}^5 ln(P(x_i))$. Then solve for $frac{dM}{dp}=0$.
- $M = sum_{i=1}^5 ln((1-p)^{x_i - 1}p)=15ln(1-p)+ 5ln p$
- $frac{dM}{dp}=0=frac{15}{1-p}+frac{5}{p} to p=-frac{1}{2}$
This is obviously wrong, as $p$ is a probability. I did this twice, what am I doing wrong?
maximum-likelihood
maximum-likelihood
asked Jan 8 at 9:39
an apprenticean apprentice
978
asked Jan 8 at 9:39
an apprenticean apprentice
978
asked Jan 8 at 9:39
an apprenticean apprentice
978
asked Jan 8 at 9:39
an apprenticean apprentice
978
asked Jan 8 at 9:39
an apprenticean apprentice
978
978
add a comment |
add a comment |
2 Answers
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From chain rule, when we differentiate, you miss out a negative sign.
It is suppose to be
$$-frac{15}{1-p}+frac5p=0$$
answered Jan 8 at 9:44
Siong Thye GohSiong Thye Goh
101k1466117
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add a comment |
$begingroup$
You may also proceed directly:
$$f(p)=prod_{i=1}^5P(x_i) = p^5(1-p)^{3+7+5+3+2-5}= p^5(1-p)^{15} stackrel{!}{rightarrow}mbox{Max for } p in (0,1)$$
You get
$$f'(p) = 0 Leftrightarrow 5(1-p)-15p = 0 Leftrightarrow boxed{p = frac{1}{4}}$$
answered Jan 8 at 10:01
trancelocationtrancelocation
10.6k1722
$endgroup$
$begingroup$
Power of the $1-p$ should be 15, not 20. Is not it? So $p=1/4$.
$endgroup$
– an apprentice
Jan 8 at 11:31
$begingroup$
OOOps. Yes. Editing correspondingly. Thanks a lot.
$endgroup$
– trancelocation
Jan 8 at 11:33
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From chain rule, when we differentiate, you miss out a negative sign.
It is suppose to be
$$-frac{15}{1-p}+frac5p=0$$
answered Jan 8 at 9:44
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
From chain rule, when we differentiate, you miss out a negative sign.
It is suppose to be
$$-frac{15}{1-p}+frac5p=0$$
answered Jan 8 at 9:44
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
From chain rule, when we differentiate, you miss out a negative sign.
It is suppose to be
$$-frac{15}{1-p}+frac5p=0$$
answered Jan 8 at 9:44
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
From chain rule, when we differentiate, you miss out a negative sign.
It is suppose to be
$$-frac{15}{1-p}+frac5p=0$$
answered Jan 8 at 9:44
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 8 at 9:44
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 8 at 9:44
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 8 at 9:44
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
add a comment |
add a comment |
$begingroup$
You may also proceed directly:
$$f(p)=prod_{i=1}^5P(x_i) = p^5(1-p)^{3+7+5+3+2-5}= p^5(1-p)^{15} stackrel{!}{rightarrow}mbox{Max for } p in (0,1)$$
You get
$$f'(p) = 0 Leftrightarrow 5(1-p)-15p = 0 Leftrightarrow boxed{p = frac{1}{4}}$$
answered Jan 8 at 10:01
trancelocationtrancelocation
10.6k1722
$endgroup$
$begingroup$
Power of the $1-p$ should be 15, not 20. Is not it? So $p=1/4$.
$endgroup$
– an apprentice
Jan 8 at 11:31
$begingroup$
OOOps. Yes. Editing correspondingly. Thanks a lot.
$endgroup$
– trancelocation
Jan 8 at 11:33
add a comment |
$begingroup$
You may also proceed directly:
$$f(p)=prod_{i=1}^5P(x_i) = p^5(1-p)^{3+7+5+3+2-5}= p^5(1-p)^{15} stackrel{!}{rightarrow}mbox{Max for } p in (0,1)$$
You get
$$f'(p) = 0 Leftrightarrow 5(1-p)-15p = 0 Leftrightarrow boxed{p = frac{1}{4}}$$
answered Jan 8 at 10:01
trancelocationtrancelocation
10.6k1722
$endgroup$
$begingroup$
Power of the $1-p$ should be 15, not 20. Is not it? So $p=1/4$.
$endgroup$
– an apprentice
Jan 8 at 11:31
$begingroup$
OOOps. Yes. Editing correspondingly. Thanks a lot.
$endgroup$
– trancelocation
Jan 8 at 11:33
add a comment |
$begingroup$
You may also proceed directly:
$$f(p)=prod_{i=1}^5P(x_i) = p^5(1-p)^{3+7+5+3+2-5}= p^5(1-p)^{15} stackrel{!}{rightarrow}mbox{Max for } p in (0,1)$$
You get
$$f'(p) = 0 Leftrightarrow 5(1-p)-15p = 0 Leftrightarrow boxed{p = frac{1}{4}}$$
answered Jan 8 at 10:01
trancelocationtrancelocation
10.6k1722
$endgroup$
You may also proceed directly:
$$f(p)=prod_{i=1}^5P(x_i) = p^5(1-p)^{3+7+5+3+2-5}= p^5(1-p)^{15} stackrel{!}{rightarrow}mbox{Max for } p in (0,1)$$
You get
$$f'(p) = 0 Leftrightarrow 5(1-p)-15p = 0 Leftrightarrow boxed{p = frac{1}{4}}$$
answered Jan 8 at 10:01
trancelocationtrancelocation
10.6k1722
edited Jan 8 at 11:34
answered Jan 8 at 10:01
trancelocationtrancelocation
10.6k1722
answered Jan 8 at 10:01
trancelocationtrancelocation
10.6k1722
answered Jan 8 at 10:01
trancelocationtrancelocation
10.6k1722
10.6k1722
$begingroup$
Power of the $1-p$ should be 15, not 20. Is not it? So $p=1/4$.
$endgroup$
– an apprentice
Jan 8 at 11:31
$begingroup$
OOOps. Yes. Editing correspondingly. Thanks a lot.
$endgroup$
– trancelocation
Jan 8 at 11:33
add a comment |
$begingroup$
Power of the $1-p$ should be 15, not 20. Is not it? So $p=1/4$.
$endgroup$
– an apprentice
Jan 8 at 11:31
$begingroup$
OOOps. Yes. Editing correspondingly. Thanks a lot.
$endgroup$
– trancelocation
Jan 8 at 11:33
$begingroup$
Power of the $1-p$ should be 15, not 20. Is not it? So $p=1/4$.
$endgroup$
– an apprentice
Jan 8 at 11:31
$begingroup$
Power of the $1-p$ should be 15, not 20. Is not it? So $p=1/4$.
$endgroup$
– an apprentice
Jan 8 at 11:31
$begingroup$
OOOps. Yes. Editing correspondingly. Thanks a lot.
$endgroup$
– trancelocation
Jan 8 at 11:33
$begingroup$
OOOps. Yes. Editing correspondingly. Thanks a lot.
$endgroup$
– trancelocation
Jan 8 at 11:33
add a comment |
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