3
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Yes, there was a typo. Sorry about that. $\endgroup$ Commented Jun 7, 2022 at 13:19
  • $\begingroup$ You might want to skip the following piec in you question which seems superfluous: Now we introduce the shifted exponential distribution: $$F(z|\lambda,z_0)=\begin{cases}0\;\;\;\;\text{ if }z<z_0\\1-e^{-\lambda(z-z_0)}\;\;\;\;\text{if }z\geq z_0\end{cases}$$ $\endgroup$ Commented Jun 7, 2022 at 17:11

2 Answers 2

4
$\begingroup$

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:

$$E[Z|z<\Delta] = \frac{\int_0^\Delta z \lambda e^{-\lambda z} dz}{\int_0^\Delta \lambda e^{-\lambda z} dz}$$

$\endgroup$
4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.