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Conditional probability of Negative Binomial R.V. given the SUM of its values
$begingroup$
Suppose ${z_{ij}}$ are independent Negative Binomial random variables with means ${mu_{ij}}$, with $i=1dots I$ and $j=1dots J$.
How do you find the (expectation of) conditional probability function of $z_{ij}$ given that $y_i = sum_{j=1}^{J} z_{ij}$ ?
I know that applying the definition of conditional probability you can get the following:
begin{align}
Pbigg[ z_{i1} = k bigg| sum_j z_{ij} = y_ibigg] &= frac{P[ z_{i1} = k ∩ sum_{j=1}^{J} z_{ij} = y_i]}{P[ sum_{j=1}^{J} z_{ij} = y_i]} \
&= frac{P[ z_{i1} = k ∩ sum_{j=2}^{J} z_{ij} = y_i - k]}{P[ sum_{j=1}^{J} z_{ij} = y_i]} \
&= frac{P[ z_{i1} = k] P[sum_{j=2}^{J} z_{ij} = y_i - k]}{P[ sum_{j=1}^{J} z_{ij} = y_i]}
end{align}
Now, IF we where dealing with Poisson R.V.s (which are quite close relatives of NB ones), we could plug into the last equality the following distributions:
$$
z_{i1} sim Poissbigg(mu_{ij}bigg)
$$
$$
sum_{j=2}^{J} z_{ij} sim Poissbigg(sum_{j=2}^{J} mu_{ij}bigg)
$$
$$
sum_{j=1}^{J} z_{ij} sim Poissbigg(sum_{j=1}^{J} mu_{ij}bigg)
$$
and this would result in:
begin{align}
Pbigg[z_{i1} bigg| sum_j z_{ij} = y_i bigg] &sim Binomialbigg(sum_j z_{ij}, frac{mathbb{E}(z_{i1})}{sum_jmathbb{E}(z_{i1})} bigg) \
&sim Binomialbigg(y_{i}, frac{mu_{i1}}{sum_jmu_{ij}} bigg)
end{align}
and the expectation of a Binomial distr $Bin(alpha, beta)$ is equal to $alpha beta$,
$$
mathbb{E}_{ P[z_{ij} | sum_j z_{ij} = y_i ] } [z_{i1}]= y_i * frac{mu_{i1}}{sum_jmu_{ij}}
$$
How can I achieve a similar result when dealing with Negative Binomial random variables?
conditional-expectation conditional-probability poisson-distribution negative-binomial
asked Jan 18 at 14:30
mscipiomscipio
11
$endgroup$
add a comment |
$begingroup$
Suppose ${z_{ij}}$ are independent Negative Binomial random variables with means ${mu_{ij}}$, with $i=1dots I$ and $j=1dots J$.
How do you find the (expectation of) conditional probability function of $z_{ij}$ given that $y_i = sum_{j=1}^{J} z_{ij}$ ?
I know that applying the definition of conditional probability you can get the following:
begin{align}
Pbigg[ z_{i1} = k bigg| sum_j z_{ij} = y_ibigg] &= frac{P[ z_{i1} = k ∩ sum_{j=1}^{J} z_{ij} = y_i]}{P[ sum_{j=1}^{J} z_{ij} = y_i]} \
&= frac{P[ z_{i1} = k ∩ sum_{j=2}^{J} z_{ij} = y_i - k]}{P[ sum_{j=1}^{J} z_{ij} = y_i]} \
&= frac{P[ z_{i1} = k] P[sum_{j=2}^{J} z_{ij} = y_i - k]}{P[ sum_{j=1}^{J} z_{ij} = y_i]}
end{align}
Now, IF we where dealing with Poisson R.V.s (which are quite close relatives of NB ones), we could plug into the last equality the following distributions:
$$
z_{i1} sim Poissbigg(mu_{ij}bigg)
$$
$$
sum_{j=2}^{J} z_{ij} sim Poissbigg(sum_{j=2}^{J} mu_{ij}bigg)
$$
$$
sum_{j=1}^{J} z_{ij} sim Poissbigg(sum_{j=1}^{J} mu_{ij}bigg)
$$
and this would result in:
begin{align}
Pbigg[z_{i1} bigg| sum_j z_{ij} = y_i bigg] &sim Binomialbigg(sum_j z_{ij}, frac{mathbb{E}(z_{i1})}{sum_jmathbb{E}(z_{i1})} bigg) \
&sim Binomialbigg(y_{i}, frac{mu_{i1}}{sum_jmu_{ij}} bigg)
end{align}
and the expectation of a Binomial distr $Bin(alpha, beta)$ is equal to $alpha beta$,
$$
mathbb{E}_{ P[z_{ij} | sum_j z_{ij} = y_i ] } [z_{i1}]= y_i * frac{mu_{i1}}{sum_jmu_{ij}}
$$
How can I achieve a similar result when dealing with Negative Binomial random variables?
conditional-expectation conditional-probability poisson-distribution negative-binomial
asked Jan 18 at 14:30
mscipiomscipio
11
$endgroup$
add a comment |
$begingroup$
Suppose ${z_{ij}}$ are independent Negative Binomial random variables with means ${mu_{ij}}$, with $i=1dots I$ and $j=1dots J$.
How do you find the (expectation of) conditional probability function of $z_{ij}$ given that $y_i = sum_{j=1}^{J} z_{ij}$ ?
I know that applying the definition of conditional probability you can get the following:
begin{align}
Pbigg[ z_{i1} = k bigg| sum_j z_{ij} = y_ibigg] &= frac{P[ z_{i1} = k ∩ sum_{j=1}^{J} z_{ij} = y_i]}{P[ sum_{j=1}^{J} z_{ij} = y_i]} \
&= frac{P[ z_{i1} = k ∩ sum_{j=2}^{J} z_{ij} = y_i - k]}{P[ sum_{j=1}^{J} z_{ij} = y_i]} \
&= frac{P[ z_{i1} = k] P[sum_{j=2}^{J} z_{ij} = y_i - k]}{P[ sum_{j=1}^{J} z_{ij} = y_i]}
end{align}
Now, IF we where dealing with Poisson R.V.s (which are quite close relatives of NB ones), we could plug into the last equality the following distributions:
$$
z_{i1} sim Poissbigg(mu_{ij}bigg)
$$
$$
sum_{j=2}^{J} z_{ij} sim Poissbigg(sum_{j=2}^{J} mu_{ij}bigg)
$$
$$
sum_{j=1}^{J} z_{ij} sim Poissbigg(sum_{j=1}^{J} mu_{ij}bigg)
$$
and this would result in:
begin{align}
Pbigg[z_{i1} bigg| sum_j z_{ij} = y_i bigg] &sim Binomialbigg(sum_j z_{ij}, frac{mathbb{E}(z_{i1})}{sum_jmathbb{E}(z_{i1})} bigg) \
&sim Binomialbigg(y_{i}, frac{mu_{i1}}{sum_jmu_{ij}} bigg)
end{align}
and the expectation of a Binomial distr $Bin(alpha, beta)$ is equal to $alpha beta$,
$$
mathbb{E}_{ P[z_{ij} | sum_j z_{ij} = y_i ] } [z_{i1}]= y_i * frac{mu_{i1}}{sum_jmu_{ij}}
$$
How can I achieve a similar result when dealing with Negative Binomial random variables?
conditional-expectation conditional-probability poisson-distribution negative-binomial
asked Jan 18 at 14:30
mscipiomscipio
11
$endgroup$
Suppose ${z_{ij}}$ are independent Negative Binomial random variables with means ${mu_{ij}}$, with $i=1dots I$ and $j=1dots J$.
How do you find the (expectation of) conditional probability function of $z_{ij}$ given that $y_i = sum_{j=1}^{J} z_{ij}$ ?
I know that applying the definition of conditional probability you can get the following:
begin{align}
Pbigg[ z_{i1} = k bigg| sum_j z_{ij} = y_ibigg] &= frac{P[ z_{i1} = k ∩ sum_{j=1}^{J} z_{ij} = y_i]}{P[ sum_{j=1}^{J} z_{ij} = y_i]} \
&= frac{P[ z_{i1} = k ∩ sum_{j=2}^{J} z_{ij} = y_i - k]}{P[ sum_{j=1}^{J} z_{ij} = y_i]} \
&= frac{P[ z_{i1} = k] P[sum_{j=2}^{J} z_{ij} = y_i - k]}{P[ sum_{j=1}^{J} z_{ij} = y_i]}
end{align}
Now, IF we where dealing with Poisson R.V.s (which are quite close relatives of NB ones), we could plug into the last equality the following distributions:
$$
z_{i1} sim Poissbigg(mu_{ij}bigg)
$$
$$
sum_{j=2}^{J} z_{ij} sim Poissbigg(sum_{j=2}^{J} mu_{ij}bigg)
$$
$$
sum_{j=1}^{J} z_{ij} sim Poissbigg(sum_{j=1}^{J} mu_{ij}bigg)
$$
and this would result in:
begin{align}
Pbigg[z_{i1} bigg| sum_j z_{ij} = y_i bigg] &sim Binomialbigg(sum_j z_{ij}, frac{mathbb{E}(z_{i1})}{sum_jmathbb{E}(z_{i1})} bigg) \
&sim Binomialbigg(y_{i}, frac{mu_{i1}}{sum_jmu_{ij}} bigg)
end{align}
and the expectation of a Binomial distr $Bin(alpha, beta)$ is equal to $alpha beta$,
$$
mathbb{E}_{ P[z_{ij} | sum_j z_{ij} = y_i ] } [z_{i1}]= y_i * frac{mu_{i1}}{sum_jmu_{ij}}
$$
How can I achieve a similar result when dealing with Negative Binomial random variables?
conditional-expectation conditional-probability poisson-distribution negative-binomial
conditional-expectation conditional-probability poisson-distribution negative-binomial
asked Jan 18 at 14:30
mscipiomscipio
11
asked Jan 18 at 14:30
mscipiomscipio
11
asked Jan 18 at 14:30
mscipiomscipio
11
asked Jan 18 at 14:30
mscipiomscipio
11
asked Jan 18 at 14:30
mscipiomscipio
11
11
add a comment |
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