Exponential distribution

[kojiT]

Q. Random samples X1, X2, X3, X4 are selected from Exponential population E(1). When the minimum of them is X, which means X= min{X1, X2, X3, X4}, what is P(X >= 0.5)? (e is allowed to use)

I have learned Exponential distribution, Moment generating function, and central limit theorem. Yet, I could not solve this question. Does any one have any idea of the way to solve this question?

 

[BigBeachBanana]

Q. Random samples X1, X2, X3, X4 are selected from Exponential population E(1). When the minimum of them is X, which means X= min{X1, X2, X3, X4}, what is P(X >= 0.5)? (e is allowed to use)

I have learned Exponential distribution, Moment generating function, and central limit theorem. Yet, I could not solve this question. Does any one have any idea of the way to solve this question?

Have you learned Order Statistics?

 

[kojiT]

Thank you for cooperation.
I just looked up order statistics, but I do not think I learned it.

random sample average or variance might be used, but I am not sure...
_ _
E[?] = 1/?, V[?] = 1/n?^2

 

[BigBeachBanana]

Thank you for cooperation.
I just looked up order statistics, but I do not think I learned it.

random sample average or variance might be used, but I am not sure...
_ _
E[?] = 1/?, V[?] = 1/n?^2

Since you don't know anything about it, we can develop intuition about how it came about. Let's do some critical thinking.
If the [imath]\text{min}(X_1,...,X_5) \ge 0.5[/imath], what does that tell you about each [imath]X_i?[/imath]

 

[kojiT]

Each of them should be Xi≥0.5.
If the question is the p(0.5<= Ex(1)), the answer should be "e^(-0.5)".

I am not sure how I can incorporate minimum Xi.
Hopefully, this makes sense.

 

[BigBeachBanana]

Each of them should be Xi≥0.5.

Correct. Furthermore, by assuming they are i.i.d's...
[math]\Pr(\text{min}(X_1,...,X_5) \ge 0.5) =\\ \Pr(X_1 \ge 0.5,...,X_5 \ge 0.5) \stackrel{i.i.d}{=}\\ \Pr(X_1 \ge 0.5)\times...\times\Pr(X_5 \ge 0.5) \stackrel{i.i.d}{=} \\ \Pr(X_i \ge 0.5)^5[/math]

 

[kojiT]

Correction
p(0.5<= X)

Given each sample is independent, p can be multiplied four times, so the answer should be e^(-2).

 

[kojiT]

Correct. Furthermore, by assuming they are i.i.d's...
[math]\Pr(\text{min}(X_1,...,X_5) \ge 0.5) =\\ \Pr(X_1 \ge 0.5,...,X_5 \ge 0.5) \stackrel{i.i.d}{=}\\ \Pr(X_1 \ge 0.5)\times...\times\Pr(X_5 \ge 0.5) \stackrel{i.i.d}{=} \\ \Pr(X_i \ge 0.5)^5[/math]

Thank you! I completely missed the fact that each sample is independent. So the answer should be e^(-2).

 

[BigBeachBanana]

Thank you! I completely missed the fact that each sample is independent. So the answer should be e^(-2).

If [imath]X_i \sim \text{Exp}(\lambda = 1)[/imath], then [imath]\Pr(X_i \ge 0.5) = e^{\frac{-.5}{1}}[/imath]

It follows that [imath]\Pr(X_i \ge 0.5)^4 = (e^{-.5})^4=e^{-2}[/imath]

Edited

 

[kojiT]

If [imath]X_i \sim \text{Exp}(\lambda = 1)[/imath], then [imath]\Pr(X_i \ge 0.5) = e^{\frac{-.5}{1}}.[/imath]

It follows that [imath]\Pr(X_i \ge 0.5)^5 = (e^{-.5})^5=e^{-2.5}.[/imath]

I really appreciate your cooperation, but the samples are {X1, X2, X3, X4}, so I think e^(-2) is the answer. Do I need to add 1 to it?

 

[BigBeachBanana]

I really appreciate your cooperation, but the samples are {X1, X2, X3, X4}, so I think e^(-2) is the answer. Do I need to add 1 to it?

My mistake. I thought there were 5. You're correct.

PS: Don't tell anyone. They'll put me back in the corner. It's dark and hot there.

 

[BigBeachBanana]

@kojiT If you're interested in the general case look into Order Statistics.
It also discusses [imath]\Pr(\text{max}\{X_1,...,X_n\} \le x)[/imath]--Similar idea.
As well as the joint distribution [imath]\Pr(\text{min}\{X_1,...,X_n\} \le x,\text{max}\{X_1,...,X_n\} \le y )[/imath].