2
$\begingroup$

We are given $(X_1, X_2, X_3, Y_1, Y_2)$ where $X_1 \sim \text{Exp}(1)$, $X_2 \sim \text{Exp}(2)$, $X_3 \sim \text{Exp}(3)$, $Y_1 \sim\text{Exp}(4)$, $Y_2 \sim \text{Exp}(4)$ and are asked the probability that the minimum of $X_1, X_2, X_3$ is less than the maximum of $Y_1$ and $Y_2$.

I found that the distribution of $\min(X_1, X_2, X_3)$ is $\text{Exp}(6)$ and the distribution of $\max(Y_1, Y_2)$ is $\text{Exp}(4)+\text{Exp}(8)$.

My question is then how do I compute the probability that $\text{Exp}(6) < \text{Exp}(4) + \text{Exp}(8)$

$\endgroup$
2
  • $\begingroup$ Are the variables all mutually independents? $\endgroup$
    – Frostic
    Apr 27, 2018 at 18:27
  • $\begingroup$ Yes extracharacters $\endgroup$
    – GTOgod
    Apr 27, 2018 at 18:28

2 Answers 2

Reset to default
3
$\begingroup$

One wants to compute $$p=P(\min(X_1,X_2,X_3)<\max(Y_1,Y_2))=1-P(\min(X_1,X_2,X_3)>\max(Y_1,Y_2))$$ To do so, first note that, for every nonnegative $y$, $$P(\min(X_1,X_2,X_3)>y)=P(X_1>y)P(X_2>y)(X_3>y)=e^{-y}\cdot e^{-2y}\cdot e^{-3y}=e^{-6y}$$ hence, conditioning on $\max(Y_1,Y_2)$, $$1-p=E(P(\min(X_1,X_2,X_3)>\max(Y_1,Y_2)\mid\max(Y_1,Y_2))=E(e^{-6\max(Y_1,Y_2)})$$ Now, for every random variable $Z$ such that $0<Z<1$ almost surely, $$E(Z)=\int_0^1P(Z>z)dz$$ hence $$E(e^{-6\max(Y_1,Y_2)})=\int_0^1 P(e^{-6\max(Y_1,Y_2)}>z)dz=\int_0^1P(\max(Y_1,Y_2)<-\tfrac16\ln z)dz$$ Furthermore, for every nonnegative $y$, $$P(\max(Y_1,Y_2)<y)=P(Y_1<y)\cdot P(Y_2<y)=(1-e^{-4y})^2$$ hence $$1-p=\int_0^1(1-z^{2/3})^2dz\stackrel{z^2=t^3}=\int_0^1(1-t)^2\frac32t^{1/2}dt=\frac32\frac{\Gamma(3)\Gamma(3/2)}{\Gamma(9/2)}$$ that is, $$p=1-\frac32\frac{2}{(7/2)(5/2)(3/2)}=1-\frac8{35}=\frac{27}{35}$$

$\endgroup$
3
  • $\begingroup$ I have been trying to figure out how you get the line after "hence, conditioning on $\max(Y_1,Y_2)$". Would you please help me understand it? $\endgroup$
    – Shashi
    Apr 27, 2018 at 20:19
  • 1
    $\begingroup$ @Shashi This is simply the fact that, for every event $A$ and every random variable $Z$, $$P(A)=E(P(A\mid Z))$$ which is itself a special case of the fact that, for every integrable random variable $W$ and every random variable $Z$, $$E(W)=E(E(W\mid Z))$$ which is itself a special case of the fact that, for every integrable random variable $W$ and every event $A$ in some sub-sigma-algebra $\mathcal G$ of the sigma-algebra defining $P$, $$E(W\mathbf 1_A)=E(E(W\mid \mathcal G)\mathbf 1_A)$$ which is actually part of the very definition of the random variable $$E(W\mid \mathcal G)$$ $\endgroup$
    – Did
    Apr 27, 2018 at 20:53
  • $\begingroup$ Many thanks! I understand it now. $\endgroup$
    – Shashi
    Apr 27, 2018 at 21:06
0
$\begingroup$

$U = \min(X_1,X_2,X_3)$

$\forall u>0, \ P(U>u) = P(X_1>u)P(X_2>u)P(X_3>u) = e^{-u}e^{-2u}e^{-3u}=e^{-6u}$

$U\sim \mathcal E (6)$

$V = \max(Y_1,Y_2)$

$\forall v>0, \ P(V<v) = P(Y_1<v)P(Y_2<v) = (1-e^{-4u})^2$

$\forall v>0, \ f_V(v) = 8e^{-4u}(1-e^{-4u})$

Using the law of total probability $$\begin{align} P(U<V)& = \int_{0}^{\infty} P(U<v) \ f_V(v) \ dv \\ &= \int_0^{\infty} (1-e^{- 6v}) \ 8(e^{-4v}-e^{-8v}) \ dv \\ &= \int_0^{\infty} 8(e^{-4v} - e^{-8v} - e^{-10v} + e^{14v}) \ dv \\ &= 2-1-\frac 8 {10}+\frac 8 {14} \\ &= 1- \frac 45+ \frac 47 = 1 - \frac{8}{35} = = \frac{27}{35}\end{align}$$

$\endgroup$
6
  • $\begingroup$ This makes sense to me, but when I compute this I get 0.84. My professors answer is $\frac{27}{35}$, which he got by integrating $\int_{0}^{\infty} 6e^{-6x}(2e^{-4x} -e^{-8x})$. Any idea where he could have generated this from? $\endgroup$
    – GTOgod
    Apr 27, 2018 at 18:34
  • $\begingroup$ I added how to compute the probability $P(Z>Y_1)$. Following the same logic you can compute $P(U<Y+Z)$ with $U \sim \mathcal E (6)$, $Y \sim \mathcal E (4)$ and $Y \sim \mathcal E (8)$. Just apply the law of total probability twice. Let me know if you do not get the same formula as your teacher $\endgroup$
    – Frostic
    Apr 27, 2018 at 19:12
  • $\begingroup$ Which parts in the revised version, the final "numerical" result at the end, are not directly copied on/inspired by another answer? $\endgroup$
    – Did
    Apr 28, 2018 at 6:40
  • $\begingroup$ My answer is way cleaner and totally different from yours $\endgroup$
    – Frostic
    Apr 28, 2018 at 9:51
  • $\begingroup$ Would you be delusional? (Anyway, please use @.) $\endgroup$
    – Did
    Apr 29, 2018 at 4:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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