Distribution of Exponential Random Variable (RV)

In summary: The factor of \lambda outside the integral comes from the differential element, dx, when you make the substitution x-->\lambdax, as gel suggested. Sorry I didn't notice the equivalence of the expressions yesterday. I was distracted by women's iceskating on TV. :blushing:
  • #1
EngWiPy
1,368
61
Hello,

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

Thanks in advance.
 
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  • #2
saeddawoud said:
Hello,

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

Thanks in advance.


I'm pretty sure it is

[tex]f_{X_1}(x)=\lambda e^{-\lambda x}[/tex]

You can check that this distribution has mean [tex]1/\lambda[/tex]. If [tex]\lambda[/tex] is the hazard rate, then [tex]\mu = 1/\lambda[/tex] is the mean waiting time for the process to end (for example, X could be the lifetime of a battery).
 
  • #3
Adeimantus said:
I'm pretty sure it is

[tex]f_{X_1}(x)=\lambda e^{-\lambda x}[/tex]

You can check that this distribution has mean [tex]1/\lambda[/tex]. If [tex]\lambda[/tex] is the hazard rate, then [tex]\mu = 1/\lambda[/tex] 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 [tex]X_1~exp(\lambda)[/tex] and [tex]X_2~exp(\mu)[/tex]. Then, based on that, defined a new RV say [tex]Z=\frac{X_1X_2}{X_1+X_2+1}[/tex], and found the Commulative Distribtion Function (CDF) of [tex]Z[/tex] as:

[tex]
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
[/tex]
I tried to do it by myself and got another result which is:
[tex]1-\mu e^{-z(\lambda+\mu)}\int_0^\infty exp \left[\frac{-\lambda z(z+1)}{x}-\mu x\right] \,dx[/tex]

Did I do something wrong in my derivation?
 
Last edited:
  • #4
saeddawoud said:
Ok, fine. Now, let me elaborate about the point where I had stopped. The authers defined two exponentially distributed RVs [tex]X_1~exp(\lambda)[/tex] and [tex]X_2~exp(\mu)[/tex]. Then, based on that, defined a new RV say [tex]Z=\frac{X_1X_2}{X_1+X_2+1}[/tex], and found the Commulative Distribtion Function (CDF) of [tex]Z[/tex] as:

[tex]
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 r^{-z(\lambda+\mu)} \int_0^\infty exp\left[\frac{-z(z+1)}{x}-\lambda\mu x\right] \, dx
[/tex]
I tried to do it by myself and got another result which is:
[tex]1-\mu e^{-z(\lambda+\mu)}\int_0^\infty exp \left[\frac{-\lambda z(z+1)}{x}-\mu x\right] \,dx[/tex]

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?
 
  • #5
Adeimantus said:
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:

[tex]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 [/tex]

[tex]=\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 [/tex]

But:

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

Then:

[tex]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 [/tex]

[tex]=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[/tex]

Now, multiply [tex]f_{X_2}(\beta)[/tex] by the terms in the cirular brackets inside the integral, and then make change of variables [tex]x = \beta-z[/tex], and after some simple mathematical manipulations you get my answer.
 
  • #6
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:
  • #7
saeddawoud said:
[tex]
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
[/tex]
I tried to do it by myself and got another result which is:
[tex]1-\mu e^{-z(\lambda+\mu)}\int_0^\infty exp \left[\frac{-\lambda z(z+1)}{x}-\mu x\right] \,dx[/tex]

Did I do something wrong in my derivation?

Those expressions are the same. Try substituting [itex]\lambda x[/itex] for x in your expression.
 
  • #8
gel said:
Those expressions are the same. Try substituting [itex]\lambda x[/itex] for x in your expression.
In this way the expression inside the exponential function will be the same. But, what about [tex]\lambda[/tex] in the author's expression which is outside the integral? How did they get it?
 
  • #9
saeddawoud said:
In this way the expression inside the exponential function will be the same. But, what about [tex]\lambda[/tex] in the author's expression which is outside the integral? How did they get it?

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

Related to Distribution of Exponential Random Variable (RV)

What is an exponential random variable (RV)?

An exponential random variable is a type of continuous probability distribution that is commonly used to model the time between events in a Poisson process. It is characterized by a single parameter, λ (lambda), which represents the rate of occurrence of the events.

How is the distribution of an exponential RV calculated?

The probability density function (PDF) of an exponential RV is given by f(x) = λe^(-λx), where x is the time between events. The cumulative distribution function (CDF) is F(x) = 1 - e^(-λx). The mean and variance of an exponential RV are both equal to 1/λ.

What are the key properties of the exponential distribution?

The exponential distribution is a continuous distribution that is always positive, with a right-skewed shape. It has a single mode and no upper bound. The mean and median of an exponential RV are not equal, and the distribution is memoryless.

How is the exponential distribution used in real-world applications?

The exponential distribution has many practical applications, including in reliability engineering, queuing theory, and survival analysis. It is commonly used to model the time between failures of a system, the time between customer arrivals in a queue, and the time until an event occurs (e.g. time until a machine breaks down or a patient dies).

Can an exponential RV take on negative values?

No, an exponential RV cannot take on negative values. Since it is a continuous distribution, the probability of any specific value is zero. However, it is possible to have an event occur immediately (x=0) with non-zero probability.

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