What is the zero-truncated Poisson distribution used for? And how is the mean and variance derived? I know that the density looks like this: $P(Y=y) = \frac{e^{-\lambda} \lambda^y}{y!(1-e^{-\lambda})}$
and from wikipedia that the mean and variance like this: 
$$\operatorname E(Y) = \frac{\lambda}{1-e^{-\lambda}}$$
$$ \operatorname{Var}(Y) = \frac{\lambda + \lambda^2}{1-e^{- \lambda}} - \frac{\lambda^2}{(1-e^{- \lambda})^2}$$
But how have the mean and variance been derived?
 A: Note that your PMF is $p(y)=\frac{f(y)}{1-e^{-\lambda}}$, for $y>0$,  where $f(y)$ is the PMF of normal Poisson.
Then,
$$\operatorname E[Y]=\sum_{y=1}^\infty y \frac{f(y)}{1-e^{-\lambda}}=\frac{1}{1-e^{-\lambda}}\sum_{y=1}^\infty y f(y)=\frac{1}{1-e^{-\lambda}}\underbrace{\sum_{y=0}^\infty yf(y)}_{\lambda \ \ \ \text{(i.e. Poisson Mean)}}=\frac{\lambda}{1-e^{-\lambda}}$$
Similarly, we can find $E[Y^2]$ as $\frac{\lambda+\lambda^2}{1-e^{-\lambda}}$, yielding a variance equation (yours have an extra $\pm$ in the first summand):
$$\operatorname{var}(Y)=\frac{\lambda+\lambda^2}{1-e^{-\lambda}}-\frac{\lambda^2}{(1-e^{-\lambda})^2}$$
A: $\newcommand{\e}{\operatorname E}\newcommand{\v}{\operatorname{var}}$This appears to include one or more applications of the zero-truncated Poisson distribution in applied statistics.
Suppose $X\sim\operatorname{Poisson}(\lambda)$ and $Y= \begin{cases} 1 & \text{if } X=0, \\ 0 & \text{otherwise.} \end{cases}$
The zero-truncated Poisson distribution is the conditional distribution of $X$ given then event $Y=0.$ So we seek $\e(X\mid Y=0)$ and $\v(X\mid Y=0).$
First, apply the law of total expectation:
\begin{align}
& \lambda =\e(X) = \e(\e(X\mid Y)) \\[6pt]
= {} & \e(X\mid Y=0)\Pr(Y=0) + \e(X\mid Y=1)\Pr(Y=1) \\[6pt]
= {} & \e(X\mid Y=0)(1-e^{-\lambda}) + 0.
\end{align}
So we have
$$
\lambda =\e(X\mid Y=0)(1-e^{-\lambda} )
$$
and therefore
$$
\e(X\mid Y=0) = \frac \lambda {1-e^{-\lambda}},
$$
or, if you like,
$$
\e(X\mid Y=0) = \frac{\lambda e^\lambda}{e^\lambda -1}.
$$
Next apply the law of total variance:
$$
\lambda = \v(X) = \v(\e(X\mid Y)) + \e(\v(X\mid Y)). \tag 0
$$
And
\begin{align}
\e(\v(X\mid Y)) = {} & \v(X\mid Y=0)\Pr(Y=0) \\
& {} + \v(X\mid Y=1)\Pr(Y=1) \\[6pt]
= {} & \v(X\mid Y=0) (1-e^{-\lambda}) + 0. \tag 1
\end{align}
\begin{align}
\v(\e(X\mid Y)) = {} & \v\left.\begin{cases} 0 & \text{if } Y = 1, \\ \lambda/(1-e^{-\lambda}) & \text{if } Y=0 \end{cases} \right\} \\[10pt]
= {} & \Pr(Y=0)\cdot\Pr(Y=1) \cdot \left( \frac \lambda {1-e^{-\lambda}} \right)^2 \\[8pt]
= {} & (1-e^{-\lambda}) e^{-\lambda} \cdot \left( \frac \lambda {1-e^{-\lambda}} \right)^2 \tag 2
\end{align}
So the sum of the two quantities on lines $(1)$ and $(2)$ above is equal, as stated on line $(0),$ to $\lambda.$ From that you can deduce $\v(X\mid Y=0),$ the variance of the truncated Poisson distribution.
