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?

In addition to @SextusEmpiricus 's beautiful solution (+1), following is the direct computation (as also noted in the answer):
$$\begin{align}E[z|z<\Delta,\lambda]&=\int_0^\Delta z f(z|z<\Delta,\lambda)dz =\int_0^\Delta z \frac{\lambda e^{-\lambda z}}{F(\Delta|\lambda)} dz \\&=\frac{1}{F(\Delta|\lambda)}\int_0^\Delta \underbrace{z}_{u} \underbrace{\lambda e^{-\lambda z} dz}_{dv}\\&=\frac{1}{F(\Delta|\lambda)}\left(-ze^{-\lambda z}\bigg|_0^\Delta+\int_0^\Delta e^{-\lambda z} dz\right)\\&=\frac{1}{F(\Delta|\lambda)}\left(-\Delta e^{-\Delta\lambda}-\frac{1}{\lambda}e^{-\lambda z}\bigg|_0^\Delta\right) \\&=\frac{1}{F(\Delta|\lambda)}\left(-\Delta (1-F(\Delta|\lambda))-\frac{1}{\lambda} \underbrace{(e^{-\lambda \Delta}-1)}_{-F(\Delta|\lambda)}\right)\\&=\frac{-\Delta(1-F(\Delta|\lambda))}{F(\Delta|\lambda)}+\frac{1}{\lambda}\end{align}$$
