# Exponential RV

1. Jan 24, 2009

### EngWiPy

Hello,

In a paper, the authors defined an exponential Random Variable (RV) as $$X_1 \mbox{~EXP}(\lambda)$$ where $$\lambda$$ is the hazard rate. What will be the distribution of this RV:
$$f_{X_1}(x)=\lambda e^{-\lambda x}$$ or
$$f_{X_1}(x)=\frac{1}{\lambda} e^{-\frac{x}{\lambda}}$$

2. Jan 24, 2009

I'm pretty sure it is

$$f_{X_1}(x)=\lambda e^{-\lambda x}$$

You can check that this distribution has mean $$1/\lambda$$. If $$\lambda$$ is the hazard rate, then $$\mu = 1/\lambda$$ is the mean waiting time for the process to end (for example, X could be the lifetime of a battery).

3. Jan 24, 2009

### EngWiPy

Ok, fine. Now, let me elaborate about the point where I had stopped. The authers defined two exponentially distributed RVs $$X_1~exp(\lambda)$$ and $$X_2~exp(\mu)$$. Then, based on that, defined a new RV say $$Z=\frac{X_1X_2}{X_1+X_2+1}$$, and found the Commulative Distribtion Function (CDF) of $$Z$$ as:

$$F_Z(z)= Pr \left[ Z \leq z\right] \\ = Pr \left[ \frac{X_1X_2}{X_1+X_2+1}\leq z\right] =1-\lambda \mu e^{-z(\lambda+\mu)} \int_0^\infty exp\left[\frac{-z(z+1)}{x}-\lambda\mu x\right] \, dx$$
I tried to do it by myself and got another result which is:
$$1-\mu e^{-z(\lambda+\mu)}\int_0^\infty exp \left[\frac{-\lambda z(z+1)}{x}-\mu x\right] \,dx$$

Did I do something wrong in my derivation?

Last edited: Jan 24, 2009
4. Jan 24, 2009

Hmm. I'm not getting either of those answers. First of all, what is 'r' in the expression given by the authors? Also, can you show the steps of your derivation?

5. Jan 24, 2009

### EngWiPy

I am sorry, it is not 'r' but 'e'. Ok, the steps of my derivation are as the following:

$$F_Z(z)= \mbox{Pr}\left[X_1(X_2-z) \leq z(X_2+1)\right]=\int_0^\infty \mbox{Pr}\left[X_1(\beta-z) \leq z(\beta+1)\right] f_{X_2}(\beta) \, d\beta$$

$$=\int_0^z \mbox{Pr}\left[X_1(\beta-z) \leq z(\beta+1)\right] f_{X_2}(\beta) \, d\beta \, + \int_z^{\infty} \mbox{Pr}\left[X_1(\beta-z) \leq z(\beta+1)\right] f_{X_2}(\beta) \, d\beta$$

But:

$$\mbox{Pr}\left[X_1(\beta-z) \leq z(\beta+1) \right] = 1$$ for $$0\leq \beta \leq z$$

Then:

$$F_Z(z)=\int_0^z f_{X_2}(\beta) \, d\beta + \int_z^{\infty} \mbox{Pr}\left[X_1(\beta-z) \leq z(\beta+1)\right] f_{X_2}(\beta) \, d\beta$$

$$=F_{X_2}(z) + \int_z^{\infty}\left(1-\mbox{exp}\left[\frac{-\lambda z(\beta+1)}{\beta-z}\right]\right)\, f_{X_2}(\beta) \, d\beta$$

Now, multiply $$f_{X_2}(\beta)$$ by the terms in the cirular brackets inside the integral, and then make change of variables $$x = \beta-z$$, and after some simple mathematical manipulations you get my answer.

6. Jan 24, 2009

I follow your derivation, and it looks good, although I haven't yet done those last manipulations to get your answer. I'll try it and see if I get what you did.

edit: yes, I get the same answer you do. Perhaps the answer in the paper is a misprint?

Last edited: Jan 24, 2009
7. Jan 25, 2009

### gel

Those expressions are the same. Try substituting $\lambda x$ for x in your expression.

8. Jan 25, 2009

### EngWiPy

In this way the expression inside the exponential function will be the same. But, what about $$\lambda$$ in the author's expression which is outside the integral? How did they get it?

9. Jan 25, 2009

The factor of $$\lambda$$ outside the integral comes from the differential element, dx, when you make the substitution x-->$$\lambda$$x, as gel suggested. Sorry I didn't notice the equivalence of the expressions yesterday. I was distracted by women's iceskating on TV.