Mixing
Compute the maximum likelihood estimator for $θ$.
$begingroup$
Let $X_1, X_2, . . . , X_n$ be a random sample from a Bernoulli distribution with parameter $θ$. Compute the maximum likelihood estimator for $θ$.
In my opinion this is the correct way to solve it:
$L(Theta)=theta_1*(1-theta_1)*...theta_n*(1-theta_n)=theta^n*(1-theta)^n$
$l(Theta)=n*ln(theta)+n*ln(1-theta)$
$l'(theta)=frac{n}{theta}-frac{n}{1-theta}=0$, $theta=frac{1}{2}$
For the official solution it should be $frac{sum_{i=1}^nX_i}{n}$, why MLE is incorrect, where should I put the $x$ that in my solution is not present?
statistics
asked Jan 25 at 14:49
Luke MarciLuke Marci
856
$endgroup$
add a comment |
$begingroup$
Let $X_1, X_2, . . . , X_n$ be a random sample from a Bernoulli distribution with parameter $θ$. Compute the maximum likelihood estimator for $θ$.
In my opinion this is the correct way to solve it:
$L(Theta)=theta_1*(1-theta_1)*...theta_n*(1-theta_n)=theta^n*(1-theta)^n$
$l(Theta)=n*ln(theta)+n*ln(1-theta)$
$l'(theta)=frac{n}{theta}-frac{n}{1-theta}=0$, $theta=frac{1}{2}$
For the official solution it should be $frac{sum_{i=1}^nX_i}{n}$, why MLE is incorrect, where should I put the $x$ that in my solution is not present?
statistics
asked Jan 25 at 14:49
Luke MarciLuke Marci
856
$endgroup$
1
$begingroup$
bernoulli has no pdf, calculating max with derivatives is here not possible
$endgroup$
– Joey Doey
Jan 25 at 14:52
1
$begingroup$
Your likelihood formula is wrong - your formula doesn't depend on the data that you see, only $n$. When you use the correct formula you can differentiate the log likelihood to get the MLE.
$endgroup$
– Alex
Jan 25 at 15:28
$begingroup$
@Alex what do you mean?
$endgroup$
– Luke Marci
Jan 25 at 15:31
1
$begingroup$
The likelihood should depend on the data you see. E.g. suppose you just had $n = 1$ and observed $X_1 = x_1$. If that was a success ($x_1 = 1$) then the likelihood would be $theta$, and if it was a failure ($x_1 = 0$) then the likelihood would be $(1 - theta)$. In the first case the MLE would be $hat{theta} = 1$ and in the second case the MLE would be $hat{theta} = 0$.
$endgroup$
– Alex
Jan 25 at 15:36
add a comment |
$begingroup$
Let $X_1, X_2, . . . , X_n$ be a random sample from a Bernoulli distribution with parameter $θ$. Compute the maximum likelihood estimator for $θ$.
In my opinion this is the correct way to solve it:
$L(Theta)=theta_1*(1-theta_1)*...theta_n*(1-theta_n)=theta^n*(1-theta)^n$
$l(Theta)=n*ln(theta)+n*ln(1-theta)$
$l'(theta)=frac{n}{theta}-frac{n}{1-theta}=0$, $theta=frac{1}{2}$
For the official solution it should be $frac{sum_{i=1}^nX_i}{n}$, why MLE is incorrect, where should I put the $x$ that in my solution is not present?
statistics
asked Jan 25 at 14:49
Luke MarciLuke Marci
856
$endgroup$
Let $X_1, X_2, . . . , X_n$ be a random sample from a Bernoulli distribution with parameter $θ$. Compute the maximum likelihood estimator for $θ$.
In my opinion this is the correct way to solve it:
$L(Theta)=theta_1*(1-theta_1)*...theta_n*(1-theta_n)=theta^n*(1-theta)^n$
$l(Theta)=n*ln(theta)+n*ln(1-theta)$
$l'(theta)=frac{n}{theta}-frac{n}{1-theta}=0$, $theta=frac{1}{2}$
For the official solution it should be $frac{sum_{i=1}^nX_i}{n}$, why MLE is incorrect, where should I put the $x$ that in my solution is not present?
statistics
statistics
asked Jan 25 at 14:49
Luke MarciLuke Marci
856
asked Jan 25 at 14:49
Luke MarciLuke Marci
856
asked Jan 25 at 14:49
Luke MarciLuke Marci
856
asked Jan 25 at 14:49
Luke MarciLuke Marci
856
asked Jan 25 at 14:49
Luke MarciLuke Marci
856
856
1
$begingroup$
bernoulli has no pdf, calculating max with derivatives is here not possible
$endgroup$
– Joey Doey
Jan 25 at 14:52
1
$begingroup$
Your likelihood formula is wrong - your formula doesn't depend on the data that you see, only $n$. When you use the correct formula you can differentiate the log likelihood to get the MLE.
$endgroup$
– Alex
Jan 25 at 15:28
$begingroup$
@Alex what do you mean?
$endgroup$
– Luke Marci
Jan 25 at 15:31
1
$begingroup$
The likelihood should depend on the data you see. E.g. suppose you just had $n = 1$ and observed $X_1 = x_1$. If that was a success ($x_1 = 1$) then the likelihood would be $theta$, and if it was a failure ($x_1 = 0$) then the likelihood would be $(1 - theta)$. In the first case the MLE would be $hat{theta} = 1$ and in the second case the MLE would be $hat{theta} = 0$.
$endgroup$
– Alex
Jan 25 at 15:36
add a comment |
1
$begingroup$
bernoulli has no pdf, calculating max with derivatives is here not possible
$endgroup$
– Joey Doey
Jan 25 at 14:52
1
$begingroup$
Your likelihood formula is wrong - your formula doesn't depend on the data that you see, only $n$. When you use the correct formula you can differentiate the log likelihood to get the MLE.
$endgroup$
– Alex
Jan 25 at 15:28
$begingroup$
@Alex what do you mean?
$endgroup$
– Luke Marci
Jan 25 at 15:31
1
$begingroup$
The likelihood should depend on the data you see. E.g. suppose you just had $n = 1$ and observed $X_1 = x_1$. If that was a success ($x_1 = 1$) then the likelihood would be $theta$, and if it was a failure ($x_1 = 0$) then the likelihood would be $(1 - theta)$. In the first case the MLE would be $hat{theta} = 1$ and in the second case the MLE would be $hat{theta} = 0$.
$endgroup$
– Alex
Jan 25 at 15:36
1
1
$begingroup$
bernoulli has no pdf, calculating max with derivatives is here not possible
$endgroup$
– Joey Doey
Jan 25 at 14:52
$begingroup$
bernoulli has no pdf, calculating max with derivatives is here not possible
$endgroup$
– Joey Doey
Jan 25 at 14:52
1
1
$begingroup$
Your likelihood formula is wrong - your formula doesn't depend on the data that you see, only $n$. When you use the correct formula you can differentiate the log likelihood to get the MLE.
$endgroup$
– Alex
Jan 25 at 15:28
$begingroup$
Your likelihood formula is wrong - your formula doesn't depend on the data that you see, only $n$. When you use the correct formula you can differentiate the log likelihood to get the MLE.
$endgroup$
– Alex
Jan 25 at 15:28
$begingroup$
@Alex what do you mean?
$endgroup$
– Luke Marci
Jan 25 at 15:31
$begingroup$
@Alex what do you mean?
$endgroup$
– Luke Marci
Jan 25 at 15:31
1
1
$begingroup$
The likelihood should depend on the data you see. E.g. suppose you just had $n = 1$ and observed $X_1 = x_1$. If that was a success ($x_1 = 1$) then the likelihood would be $theta$, and if it was a failure ($x_1 = 0$) then the likelihood would be $(1 - theta)$. In the first case the MLE would be $hat{theta} = 1$ and in the second case the MLE would be $hat{theta} = 0$.
$endgroup$
– Alex
Jan 25 at 15:36
$begingroup$
The likelihood should depend on the data you see. E.g. suppose you just had $n = 1$ and observed $X_1 = x_1$. If that was a success ($x_1 = 1$) then the likelihood would be $theta$, and if it was a failure ($x_1 = 0$) then the likelihood would be $(1 - theta)$. In the first case the MLE would be $hat{theta} = 1$ and in the second case the MLE would be $hat{theta} = 0$.
$endgroup$
– Alex
Jan 25 at 15:36
add a comment |
1 Answer
1
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$begingroup$
The likelihood function for Bernoulli can be written as follows:
$$
L(theta)=begin{cases}theta, & X_1=1cr 1-theta, & X_1=0end{cases}times begin{cases}theta, & X_2=1cr 1-theta, & X_2=0end{cases}times cdotstimesbegin{cases}theta, & X_n=1cr 1-theta, & X_n=0end{cases}
$$
Note that $sum_{i=1}^n X_i=noverline X$ calculates the number of units in a sample, and $n-noverline X$ calculates the number of zeros. In the above product, $theta$ is multiplied as many times as there are units in the sample. And $1-theta$ is multiplied as many times as there are zeros in the sample. So
$$
L(theta) = theta^{noverline X}cdot (1-theta)^{n-noverline X}.
$$
Starting from here you can calculate MLE with derivatives.
answered Jan 26 at 3:47
NChNCh
6,8753825
$endgroup$
add a comment |
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$begingroup$
The likelihood function for Bernoulli can be written as follows:
$$
L(theta)=begin{cases}theta, & X_1=1cr 1-theta, & X_1=0end{cases}times begin{cases}theta, & X_2=1cr 1-theta, & X_2=0end{cases}times cdotstimesbegin{cases}theta, & X_n=1cr 1-theta, & X_n=0end{cases}
$$
Note that $sum_{i=1}^n X_i=noverline X$ calculates the number of units in a sample, and $n-noverline X$ calculates the number of zeros. In the above product, $theta$ is multiplied as many times as there are units in the sample. And $1-theta$ is multiplied as many times as there are zeros in the sample. So
$$
L(theta) = theta^{noverline X}cdot (1-theta)^{n-noverline X}.
$$
Starting from here you can calculate MLE with derivatives.
answered Jan 26 at 3:47
NChNCh
6,8753825
$endgroup$
add a comment |
$begingroup$
The likelihood function for Bernoulli can be written as follows:
$$
L(theta)=begin{cases}theta, & X_1=1cr 1-theta, & X_1=0end{cases}times begin{cases}theta, & X_2=1cr 1-theta, & X_2=0end{cases}times cdotstimesbegin{cases}theta, & X_n=1cr 1-theta, & X_n=0end{cases}
$$
Note that $sum_{i=1}^n X_i=noverline X$ calculates the number of units in a sample, and $n-noverline X$ calculates the number of zeros. In the above product, $theta$ is multiplied as many times as there are units in the sample. And $1-theta$ is multiplied as many times as there are zeros in the sample. So
$$
L(theta) = theta^{noverline X}cdot (1-theta)^{n-noverline X}.
$$
Starting from here you can calculate MLE with derivatives.
answered Jan 26 at 3:47
NChNCh
6,8753825
$endgroup$
add a comment |
$begingroup$
The likelihood function for Bernoulli can be written as follows:
$$
L(theta)=begin{cases}theta, & X_1=1cr 1-theta, & X_1=0end{cases}times begin{cases}theta, & X_2=1cr 1-theta, & X_2=0end{cases}times cdotstimesbegin{cases}theta, & X_n=1cr 1-theta, & X_n=0end{cases}
$$
Note that $sum_{i=1}^n X_i=noverline X$ calculates the number of units in a sample, and $n-noverline X$ calculates the number of zeros. In the above product, $theta$ is multiplied as many times as there are units in the sample. And $1-theta$ is multiplied as many times as there are zeros in the sample. So
$$
L(theta) = theta^{noverline X}cdot (1-theta)^{n-noverline X}.
$$
Starting from here you can calculate MLE with derivatives.
answered Jan 26 at 3:47
NChNCh
6,8753825
$endgroup$
The likelihood function for Bernoulli can be written as follows:
$$
L(theta)=begin{cases}theta, & X_1=1cr 1-theta, & X_1=0end{cases}times begin{cases}theta, & X_2=1cr 1-theta, & X_2=0end{cases}times cdotstimesbegin{cases}theta, & X_n=1cr 1-theta, & X_n=0end{cases}
$$
Note that $sum_{i=1}^n X_i=noverline X$ calculates the number of units in a sample, and $n-noverline X$ calculates the number of zeros. In the above product, $theta$ is multiplied as many times as there are units in the sample. And $1-theta$ is multiplied as many times as there are zeros in the sample. So
$$
L(theta) = theta^{noverline X}cdot (1-theta)^{n-noverline X}.
$$
Starting from here you can calculate MLE with derivatives.
answered Jan 26 at 3:47
NChNCh
6,8753825
answered Jan 26 at 3:47
NChNCh
6,8753825
answered Jan 26 at 3:47
NChNCh
6,8753825
answered Jan 26 at 3:47
NChNCh
6,8753825
6,8753825
add a comment |
add a comment |
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$begingroup$
bernoulli has no pdf, calculating max with derivatives is here not possible
$endgroup$
– Joey Doey
Jan 25 at 14:52
1
$begingroup$
Your likelihood formula is wrong - your formula doesn't depend on the data that you see, only $n$. When you use the correct formula you can differentiate the log likelihood to get the MLE.
$endgroup$
– Alex
Jan 25 at 15:28
$begingroup$
@Alex what do you mean?
$endgroup$
– Luke Marci
Jan 25 at 15:31
1
$begingroup$
The likelihood should depend on the data you see. E.g. suppose you just had $n = 1$ and observed $X_1 = x_1$. If that was a success ($x_1 = 1$) then the likelihood would be $theta$, and if it was a failure ($x_1 = 0$) then the likelihood would be $(1 - theta)$. In the first case the MLE would be $hat{theta} = 1$ and in the second case the MLE would be $hat{theta} = 0$.
$endgroup$
– Alex
Jan 25 at 15:36