Mixing
What is the maximum likelihood of a binomial distribution?
$begingroup$
i've looked everywhere I could for an answer to this question but no luck !
If I have $X_1 .... X_n$ random variables that are independent and identically distributed such as ∀ $1 <$ $i$ $<n$, $X_i$ ~ B(n,$theta$) (binomial distribution)
I know that the likelihood is :
$P_n$($theta$,x)=$prodlimits_{i}$ $binom{n}{x_i}p^{x_i}(1-p)^{n-x_i}$
but then it seems kind of hard to calculate as product, I tried to calculate log($p_n$) but then the $x_i$! causes me problems
Can you enlighten me ? what is the max-likelihood of a binomial distribution ?
parameter-estimation maximum-likelihood
asked Feb 1 at 15:51
Blueberry Blueberry
83
$endgroup$
add a comment |
$begingroup$
i've looked everywhere I could for an answer to this question but no luck !
If I have $X_1 .... X_n$ random variables that are independent and identically distributed such as ∀ $1 <$ $i$ $<n$, $X_i$ ~ B(n,$theta$) (binomial distribution)
I know that the likelihood is :
$P_n$($theta$,x)=$prodlimits_{i}$ $binom{n}{x_i}p^{x_i}(1-p)^{n-x_i}$
but then it seems kind of hard to calculate as product, I tried to calculate log($p_n$) but then the $x_i$! causes me problems
Can you enlighten me ? what is the max-likelihood of a binomial distribution ?
parameter-estimation maximum-likelihood
asked Feb 1 at 15:51
Blueberry Blueberry
83
$endgroup$
add a comment |
$begingroup$
i've looked everywhere I could for an answer to this question but no luck !
If I have $X_1 .... X_n$ random variables that are independent and identically distributed such as ∀ $1 <$ $i$ $<n$, $X_i$ ~ B(n,$theta$) (binomial distribution)
I know that the likelihood is :
$P_n$($theta$,x)=$prodlimits_{i}$ $binom{n}{x_i}p^{x_i}(1-p)^{n-x_i}$
but then it seems kind of hard to calculate as product, I tried to calculate log($p_n$) but then the $x_i$! causes me problems
Can you enlighten me ? what is the max-likelihood of a binomial distribution ?
parameter-estimation maximum-likelihood
asked Feb 1 at 15:51
Blueberry Blueberry
83
$endgroup$
i've looked everywhere I could for an answer to this question but no luck !
If I have $X_1 .... X_n$ random variables that are independent and identically distributed such as ∀ $1 <$ $i$ $<n$, $X_i$ ~ B(n,$theta$) (binomial distribution)
I know that the likelihood is :
$P_n$($theta$,x)=$prodlimits_{i}$ $binom{n}{x_i}p^{x_i}(1-p)^{n-x_i}$
but then it seems kind of hard to calculate as product, I tried to calculate log($p_n$) but then the $x_i$! causes me problems
Can you enlighten me ? what is the max-likelihood of a binomial distribution ?
parameter-estimation maximum-likelihood
parameter-estimation maximum-likelihood
asked Feb 1 at 15:51
Blueberry Blueberry
83
asked Feb 1 at 15:51
Blueberry Blueberry
83
asked Feb 1 at 15:51
Blueberry Blueberry
83
asked Feb 1 at 15:51
Blueberry Blueberry
83
asked Feb 1 at 15:51
Blueberry Blueberry
83
83
add a comment |
add a comment |
1 Answer
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$begingroup$
Take the log-likelihood function, i.e.
$$L(p) = log prod_i binom{n}{x_i} p^{x_i}(1-p)^{n-x_i}$$
which becomes
$$L(p) = sum_i logbinom{n}{x_i} p^{x_i}(1-p)^{n-x_i}$$
even more
$$L(p) = sum_i logbinom{n}{x_i} + sum_i x_ilog p + sum_i(n-x_i)log(1-p)$$
Since you're interested in the ML estimate of $p$. let's derive with respect to $p$ and set it to zero.
$$frac{d}{dp}L(p) = frac{1}{p}sum_i x_i - frac{1}{1-p}sum_i(n-x_i) = 0$$
which gives you
$$frac{1}{1-p}sum_i(n-x_i) = frac{1}{p}sum_i x_i$$
or
$$frac{sum_i(n-x_i)}{sum_i x_i} = frac{1-p}{p} = frac{1}{p} - 1$$
Now find $p$.
$$frac{1}{p} = frac{sum_i(n-x_i) + sum_i x_i}{sum_i x_i} = frac{sum_i n}{sum_i x_i} $$
answered Feb 1 at 16:00
Ahmad BazziAhmad Bazzi
8,5212824
$endgroup$
$begingroup$
thank you, it was just a matter of concentration and calculating step by step
$endgroup$
– Blueberry
Feb 1 at 16:03
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Take the log-likelihood function, i.e.
$$L(p) = log prod_i binom{n}{x_i} p^{x_i}(1-p)^{n-x_i}$$
which becomes
$$L(p) = sum_i logbinom{n}{x_i} p^{x_i}(1-p)^{n-x_i}$$
even more
$$L(p) = sum_i logbinom{n}{x_i} + sum_i x_ilog p + sum_i(n-x_i)log(1-p)$$
Since you're interested in the ML estimate of $p$. let's derive with respect to $p$ and set it to zero.
$$frac{d}{dp}L(p) = frac{1}{p}sum_i x_i - frac{1}{1-p}sum_i(n-x_i) = 0$$
which gives you
$$frac{1}{1-p}sum_i(n-x_i) = frac{1}{p}sum_i x_i$$
or
$$frac{sum_i(n-x_i)}{sum_i x_i} = frac{1-p}{p} = frac{1}{p} - 1$$
Now find $p$.
$$frac{1}{p} = frac{sum_i(n-x_i) + sum_i x_i}{sum_i x_i} = frac{sum_i n}{sum_i x_i} $$
answered Feb 1 at 16:00
Ahmad BazziAhmad Bazzi
8,5212824
$endgroup$
$begingroup$
thank you, it was just a matter of concentration and calculating step by step
$endgroup$
– Blueberry
Feb 1 at 16:03
add a comment |
$begingroup$
Take the log-likelihood function, i.e.
$$L(p) = log prod_i binom{n}{x_i} p^{x_i}(1-p)^{n-x_i}$$
which becomes
$$L(p) = sum_i logbinom{n}{x_i} p^{x_i}(1-p)^{n-x_i}$$
even more
$$L(p) = sum_i logbinom{n}{x_i} + sum_i x_ilog p + sum_i(n-x_i)log(1-p)$$
Since you're interested in the ML estimate of $p$. let's derive with respect to $p$ and set it to zero.
$$frac{d}{dp}L(p) = frac{1}{p}sum_i x_i - frac{1}{1-p}sum_i(n-x_i) = 0$$
which gives you
$$frac{1}{1-p}sum_i(n-x_i) = frac{1}{p}sum_i x_i$$
or
$$frac{sum_i(n-x_i)}{sum_i x_i} = frac{1-p}{p} = frac{1}{p} - 1$$
Now find $p$.
$$frac{1}{p} = frac{sum_i(n-x_i) + sum_i x_i}{sum_i x_i} = frac{sum_i n}{sum_i x_i} $$
answered Feb 1 at 16:00
Ahmad BazziAhmad Bazzi
8,5212824
$endgroup$
$begingroup$
thank you, it was just a matter of concentration and calculating step by step
$endgroup$
– Blueberry
Feb 1 at 16:03
add a comment |
$begingroup$
Take the log-likelihood function, i.e.
$$L(p) = log prod_i binom{n}{x_i} p^{x_i}(1-p)^{n-x_i}$$
which becomes
$$L(p) = sum_i logbinom{n}{x_i} p^{x_i}(1-p)^{n-x_i}$$
even more
$$L(p) = sum_i logbinom{n}{x_i} + sum_i x_ilog p + sum_i(n-x_i)log(1-p)$$
Since you're interested in the ML estimate of $p$. let's derive with respect to $p$ and set it to zero.
$$frac{d}{dp}L(p) = frac{1}{p}sum_i x_i - frac{1}{1-p}sum_i(n-x_i) = 0$$
which gives you
$$frac{1}{1-p}sum_i(n-x_i) = frac{1}{p}sum_i x_i$$
or
$$frac{sum_i(n-x_i)}{sum_i x_i} = frac{1-p}{p} = frac{1}{p} - 1$$
Now find $p$.
$$frac{1}{p} = frac{sum_i(n-x_i) + sum_i x_i}{sum_i x_i} = frac{sum_i n}{sum_i x_i} $$
answered Feb 1 at 16:00
Ahmad BazziAhmad Bazzi
8,5212824
$endgroup$
Take the log-likelihood function, i.e.
$$L(p) = log prod_i binom{n}{x_i} p^{x_i}(1-p)^{n-x_i}$$
which becomes
$$L(p) = sum_i logbinom{n}{x_i} p^{x_i}(1-p)^{n-x_i}$$
even more
$$L(p) = sum_i logbinom{n}{x_i} + sum_i x_ilog p + sum_i(n-x_i)log(1-p)$$
Since you're interested in the ML estimate of $p$. let's derive with respect to $p$ and set it to zero.
$$frac{d}{dp}L(p) = frac{1}{p}sum_i x_i - frac{1}{1-p}sum_i(n-x_i) = 0$$
which gives you
$$frac{1}{1-p}sum_i(n-x_i) = frac{1}{p}sum_i x_i$$
or
$$frac{sum_i(n-x_i)}{sum_i x_i} = frac{1-p}{p} = frac{1}{p} - 1$$
Now find $p$.
$$frac{1}{p} = frac{sum_i(n-x_i) + sum_i x_i}{sum_i x_i} = frac{sum_i n}{sum_i x_i} $$
answered Feb 1 at 16:00
Ahmad BazziAhmad Bazzi
8,5212824
answered Feb 1 at 16:00
Ahmad BazziAhmad Bazzi
8,5212824
answered Feb 1 at 16:00
Ahmad BazziAhmad Bazzi
8,5212824
answered Feb 1 at 16:00
Ahmad BazziAhmad Bazzi
8,5212824
8,5212824
$begingroup$
thank you, it was just a matter of concentration and calculating step by step
$endgroup$
– Blueberry
Feb 1 at 16:03
add a comment |
$begingroup$
thank you, it was just a matter of concentration and calculating step by step
$endgroup$
– Blueberry
Feb 1 at 16:03
$begingroup$
thank you, it was just a matter of concentration and calculating step by step
$endgroup$
– Blueberry
Feb 1 at 16:03
$begingroup$
thank you, it was just a matter of concentration and calculating step by step
$endgroup$
– Blueberry
Feb 1 at 16:03
add a comment |
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