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