Truncated Binomial Distribution at X=0

Meaning of Truncation

The literal meaning of truncation is to 'shorten' or 'cut-off' or 'discard' something.

We can define the truncation of a distribution as a process which results in certain values being ‘cut-off,’ thereby resulting in a ‘shortened’ distribution.

Truncated Distributions

Let $X$ be a random variable with pmf/pdf $f(x)$ with distribution function $F(x)$ with an infinite support.

Truncated Binomial Distribution (at $X=0$)

A discrete random variable $X$ is said to have truncated binomial distribution (at $X=0$) if its probability mass function is given by

$$ \begin{equation*} P(X=x)= \left\{ \begin{array}{ll} \frac{1}{(1-q^n)}\binom{n}{x} p^x q^{n-x}, & \hbox{$x=1,2, \ldots, n$;} \\ & \hbox{$0 < p,q < 1$, $p+q=1$}\\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$

Proof

The probability mass function of binomial distribution $B(n,p)$ is

$$ \begin{equation*} P(X=x)= \left\{ \begin{array}{ll} \binom{n}{x} p^x q^{n-x}, & \hbox{$x=0,1,\ldots, n$;} \\ & \hbox{$0 < p,q < 1$, $p+q=1$}\\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$
As $P(X=x)$ is a probability mass function, $\sum_{x=0}^n P(X=x) = 1$.

$$ \begin{eqnarray*} & & \sum_{x=0}^n P(X=x) = 1 \\ &\Rightarrow & \sum_{x=0}^n \binom{n}{x} p^x q^{n-x} = 1\\ &\Rightarrow & \binom{n}{0} p^0 q^{n-0} +\sum_{x=1}^n \binom{n}{x} p^x q^{n-x} = 1\\ &\Rightarrow & q^n +\sum_{x=1}^n \binom{n}{x} p^x q^{n-x} = 1\\ &\Rightarrow & \sum_{x=1}^n \binom{n}{x} p^x q^{n-x} = 1-q^n\\ &\Rightarrow & \frac{1}{1-q^n} \sum_{x=1}^n \binom{n}{x} p^x q^{n-x} = 1\\ &\Rightarrow & \sum_{x=1}^n \frac{1}{1-q^n} \binom{n}{x} p^x q^{n-x} = 1\\ &\Rightarrow & \sum_{x=1}^n P^\prime(X=x) = 1 \end{eqnarray*} $$

where $P^\prime(X=x) = \frac{1}{(1-q^n)} \binom{n}{x} p^x q^{n-x}$, $x=1,2, \ldots n$, $0 < p, q < 1$ and $p+q=1$. The probability function $P^\prime (X=x)$ is the probability mass function of truncated binomial (at $X=0$) distribution.

Mean of Truncated Binomial distribution at $X=0$

The mean of truncated Binomial distribution at $X=0$ is

$$ \begin{aligned} E(X)=\frac{np}{1-q^n} \end{aligned} $$

Proof

$$ \begin{eqnarray*} \mu_1^\prime = E(X) &=& \sum_{x=1}^n x\cdot P(X=x)\\ &=& \sum_{x=1}^n x\cdot\frac{1}{(1-q^n)}\binom{n}{x} p^x q^{n-x} \\ &=& \frac{1}{(1-q^n)}\sum_{x=1}^n \frac{x\cdot n! }{x!(n-x)!}p^x q^{n-x} \\ &=& \frac{np}{(1-q^n)}\sum_{x=1}^n \frac{(n-1)! }{(x-1)!(n-x)!}p^{x-1} q^{n-x}. \end{eqnarray*} $$

Let $x-1=y$, so $x=1\Rightarrow y=0$ and $x=n\Rightarrow y=n-1=N$ (say). And $n-x = (n-1)-(x-1) = N-y$.

$$ \begin{eqnarray*} \mu_1^\prime &=& \frac{np}{1-q^n}\sum_{y=0}^N \frac{N!}{y!(N-y)!}p^{y} q^{N-y}\\ &=& \frac{np}{1-q^n}\sum_{y=0}^N \binom{N}{y}p^{y}q^{N-y}\\ &=& \frac{np}{1-q^n}(p+q)^N\\ &=& \frac{np}{1-q^n}. \end{eqnarray*} $$

Variance of Truncated Binomial distribution at $X=0$

Variance of Truncated Binomial distribution at $X=0$ is

$$
\begin{aligned}
V(X) &=\frac{n(n-1)p^2+np}{(1-q^n)} - \frac{n^2p^2}{(1-q^n)^2}
\end{aligned}
$$

Proof

For variance, let us calculate $E[X*(X-1)]$

$$ \begin{eqnarray*} E[X(X-1)]&=& \sum_{x=1}^n x(x-1)P(X=x)\\ &=& \sum_{x=1}^n x(x-1)\cdot\frac{1}{(1-q^n)}\binom{n}{x} p^x q^{n-x} \\ &=& \frac{1}{(1-q^n)}\sum_{x=1}^n \frac{x(x-1)\cdot n! }{x!(n-x)!}p^x q^{n-x} \\ &=& \frac{n(n-1)p^2}{(1-q^n)}\sum_{x=2}^n \frac{(n-2)! }{(x-2)!(n-x)!}p^{x-2} q^{n-x}. \end{eqnarray*} $$

Let $x-2=y$ and $n-2=N$, then if $x=2 \Rightarrow y=0$ and if $x=n\Rightarrow y=n-2=N$. And $n-x = (n-2)-(x-2) = N-y$. Hence

$$ \begin{eqnarray*} \mu_{[2]}^\prime &=& \frac{n(n-1)p^2}{(1-q^n)}\sum_{y=0}^N \frac{N! }{y!(N-y)!}p^{y} q^{N-y}\\ & = & \frac{n(n-1)p^2}{(1-q^n)}\cdot (p+q)^N\\ & = & \frac{n(n-1)p^2}{(1-q^n)}. \end{eqnarray*} $$

Therefore
$$ \begin{aligned} \mu_2^\prime &= E(X(X-1))+E(X) \\ &= \frac{n(n-1)p^2}{(1-q^n)}+\frac{np}{(1-q^n)} \end{aligned} $$
Hence variance is

$$ \begin{eqnarray*} \mu_2 &=& \mu_2^\prime - {\mu_1^\prime}^2\\ & = &\frac{n(n-1)p^2}{(1-q^n)}+\frac{np}{(1-q^n)} - \bigg(\frac{np}{(1-q^n)}\bigg)^2\\ &=&\frac{n(n-1)p^2+np}{(1-q^n)} - \frac{n^2p^2}{(1-q^n)^2}. \end{eqnarray*} $$

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.

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