Distribution of Exponential Random Variable (RV)

[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
$$f_{X_1}(x)=\frac{1}{\lambda} e^{-\frac{x}{\lambda}}$$

Thanks in advance.

 

[Adeimantus]

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
$$f_{X_1}(x)=\frac{1}{\lambda} e^{-\frac{x}{\lambda}}$$

Thanks in advance.

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).

 

[EngWiPy]

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).

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:

$$F_Z(z)= Pr \left[ Z \leq z\right] \\<br /> = Pr \left[ \frac{X_1X_2}{X_1+X_2+1}\leq z\right]<br /> =1-\lambda \mu e^{-z(\lambda+\mu)} \int_0^\infty exp\left[\frac{-z(z+1)}{x}-\lambda\mu x\right] \, dx$$
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?

 

[Adeimantus]

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:

$$F_Z(z)= Pr \left[ Z \leq z\right] \\<br /> = Pr \left[ \frac{X_1X_2}{X_1+X_2+1}\leq z\right]<br /> =1-\lambda\mu r^{-z(\lambda+\mu)} \int_0^\infty exp\left[\frac{-z(z+1)}{x}-\lambda\mu x\right] \, dx$$
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?

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?

 

[EngWiPy]

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?

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.

 

[Adeimantus]

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?

 

[gel]

$$F_Z(z)= Pr \left[ Z \leq z\right] \\<br /> = Pr \left[ \frac{X_1X_2}{X_1+X_2+1}\leq z\right]<br /> =1-\lambda \mu e^{-z(\lambda+\mu)} \int_0^\infty exp\left[\frac{-z(z+1)}{x}-\lambda\mu x\right] \, dx$$
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?

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

 

[EngWiPy]

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

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?

 

[Adeimantus]

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?

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.