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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 $\text{Exp}(6) < \text{Exp}(4) + \text{Exp}(8)$

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

2 Answers 2

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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, $$p=1-\frac32\frac{2}{(7/2)(5/2)(3/2)}=1-\frac8{35}=\frac{27}{35}$$

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  • $\begingroup$ 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
    Commented Apr 27, 2018 at 20:19
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    $\begingroup$ @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
    Commented Apr 27, 2018 at 20:53
  • $\begingroup$ Many thanks! I understand it now. $\endgroup$
    – Shashi
    Commented Apr 27, 2018 at 21:06
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$U = \min(X_1,X_2,X_3)$

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

$U\sim \mathcal E (6)$


$V = \max(Y_1,Y_2)$

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

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


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

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  • $\begingroup$ 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
    Commented 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
    Commented 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
    Commented Apr 28, 2018 at 6:40
  • $\begingroup$ My answer is way cleaner and totally different from yours $\endgroup$
    – Frostic
    Commented Apr 28, 2018 at 9:51
  • $\begingroup$ Would you be delusional? (Anyway, please use @.) $\endgroup$
    – Did
    Commented Apr 29, 2018 at 4:59

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