How to computed "truncated shifted exponential distribution"? I have a research problem to solve. For regular exponential distribution,
$$F(z|\lambda)=\begin{cases}0\;\;\;\;\text{if }z<0\\1-e^{-\lambda z}\;\;\;\;\text{ if }z\geq 0\end{cases}$$ with density $$f(z|\lambda)=\begin{cases}0\;\;\;\;\text{ if }z<0\\\lambda e^{-\lambda z}\;\;\;\;\text{if }z\geq 0\end{cases}$$
Now we introduce the shifted exponential distribution:
$$F(z|\lambda,z_0)=\begin{cases}0\;\;\;\;\text{ if }z<z_0\\1-e^{-\lambda(z-z_0)}\;\;\;\;\text{if }z\geq z_0\end{cases}$$
and for $x\geq 0$ and $\Delta\geq0$, we have
$$E(z|\lambda)=\frac{1}{\lambda}$$
$$E(z|z\geq x,\lambda)=x+\frac{1}{\lambda}$$
$$E[z|z\in[x,x+\Delta],\lambda]=x+E[z|z<\Delta,\lambda]$$
Now I want to show that
$$E[z|z<\Delta,\lambda]=\frac{1}{\lambda}-\frac{\Delta \cdot [1-F(\Delta|\lambda)]}{F(\Delta|\lambda)}$$
Could somebody be kind enough to show me the steps?

We have that
$$E[Z] = E[Z|A] \cdot P(A) + E[Z|!A] \cdot P(!A)$$
or differently written
$$ E[Z|A]  =  \frac{ E[Z] - E[Z|!A] \cdot P(!A)}{P(A)}$$
Now let the event $A$ be $z<\Delta$, then this becomes
$$ \begin{array}{}
E[Z|z<\Delta]  &=&  \frac{ E[Z] - E[Z|z \geq \Delta] \cdot P(z\geq\Delta)}{P(z<\Delta)}\\
 &=& \frac{ E[Z] - E[Z|z \geq \Delta] \cdot (1-F(\Delta))}{F(\Delta)}\\
 &=& \frac{ \lambda^{-1} - (\lambda^{-1}+\Delta) \cdot (1-F(\Delta))}{F(\Delta)}\\
 &=&  \lambda^{-1}  -  \frac{  \Delta \cdot (1-F(\Delta))}{F(\Delta)}\\
\end{array}$$

I imagine that you could also compute it straightforward like working out the below integrals:
$$E[Z|z<\Delta] = \frac{\int_0^\Delta z \lambda e^{-\lambda z} dz}{\int_0^\Delta   \lambda e^{-\lambda z} dz}$$
