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Joram Soch |
BCCN Berlin |
joram.soch@bccn-berlin.de |
2022-10-07 14:03:00 -0700 |
Marginal distribution of a conditional binomial distribution |
Probability Distributions |
Univariate discrete distributions |
Binomial distribution |
Conditional binomial |
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Wikipedia |
2022 |
Binomial distribution |
Wikipedia, the free encyclopedia |
retrieved on 2022-10-07 |
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P358 |
bin-margcond |
JoramSoch |
Theorem: Let $X$ and $Y$ be two random variables where $Y$ is binomially distributed conditional on $X$
$$ \label{eq:Y-X-dist}
Y \vert X \sim \mathrm{Bin}(X, q)
$$
and $X$ also follows a binomial distribution, but with different success frequency:
$$ \label{eq:X-dist}
X \sim \mathrm{Bin}(n, p) ; .
$$
Then, the maginal distribution of $Y$ unconditional on $X$ is again a binomial distribution:
$$ \label{eq:Y-dist}
Y \sim \mathrm{Bin}(n, pq) ; .
$$
Proof: We are interested in the probability that $Y$ equals a number $m$. According to the law of marginal probability or the law of total probability, this probability can be expressed as:
$$ \label{eq:Y-dist-s1}
\mathrm{Pr}(Y = m) = \sum_{k=0}^{\infty} \mathrm{Pr}(Y = m \vert X = k) \cdot \mathrm{Pr}(X = k) ; .
$$
Since, by definitions \eqref{eq:X-dist} and \eqref{eq:Y-X-dist}, $\mathrm{Pr}(X = k) = 0$ when $k > n$ and $\mathrm{Pr}(Y = m \vert X = k) = 0$ when $k < m$, we have:
$$ \label{eq:Y-dist-s2}
\mathrm{Pr}(Y = m) = \sum_{k=m}^{n} \mathrm{Pr}(Y = m \vert X = k) \cdot \mathrm{Pr}(X = k) ; .
$$
Now we can take the probability mass function of the binomial distribution and plug it in for the terms in the sum of \eqref{eq:Y-dist-s2} to get:
$$ \label{eq:Y-dist-s3}
\mathrm{Pr}(Y = m) = \sum_{k=m}^{n} {k \choose m} , q^m , (1-q)^{k-m} \cdot {n \choose k} , p^k , (1-p)^{n-k} ; .
$$
Applying the binomial coefficient identity ${n \choose k} {k \choose m} = {n \choose m} {n-m \choose k-m}$ and rearranging the terms, we have:
$$ \label{eq:Y-dist-s4}
\mathrm{Pr}(Y = m) = \sum_{k=m}^{n} {n \choose m} , {n-m \choose k-m} , p^k , q^m , (1-p)^{n-k} , (1-q)^{k-m} ; .
$$
Now we partition $p^k = p^m \cdot p^{k-m}$ and pull all terms dependent on $k$ out of the sum:
$$ \label{eq:Y-dist-s5}
\begin{split}
\mathrm{Pr}(Y = m) &= {n \choose m} , p^m , q^m \sum_{k=m}^{n} {n-m \choose k-m} , p^{k-m} , (1-p)^{n-k} , (1-q)^{k-m} \\
&= {n \choose m} , (p q)^m \sum_{k=m}^{n} {n-m \choose k-m} , \left( p (1-q) \right)^{k-m} , (1-p)^{n-k} ; .
\end{split}
$$
Then we subsititute $i = k - m$, such that $k = i + m$:
$$ \label{eq:Y-dist-s6}
\mathrm{Pr}(Y = m) = {n \choose m} , (p q)^m \sum_{i=0}^{n-m} {n-m \choose i} , \left( p - pq \right)^{i} , (1-p)^{n-m-i} ; .
$$
According to the binomial theorem
$$ \label{eq:bin-th}
(x+y)^n = \sum_{k=0}^{n} {n \choose k} , x^{n-k} , y^k ; ,
$$
the sum in equation \eqref{eq:Y-dist-s6} is equal to
$$ \label{eq:Y-dist-sum}
\sum_{i=0}^{n-m} {n-m \choose i} , \left( p - pq \right)^{i} , (1-p)^{n-m-i} = \left( (p-pq)+(1-p) \right)^{n-m} ; .
$$
Thus, \eqref{eq:Y-dist-s6} develops into
$$ \label{eq:Y-dist-s7}
\begin{split}
\mathrm{Pr}(Y = m) &= {n \choose m} , (p q)^m (p - pq + 1 - p)^{n-m} \\
&= {n \choose m} , (p q)^m (1 - pq)^{n-m}
\end{split}
$$
which is the probability mass function of the binomial distribution with parameters $n$ and $pq$, such that
$$ \label{eq:Y-dist-qed}
Y \sim \mathrm{Bin}(n, pq) ; .
$$