I got this problem within others as homework, and I don't know how to do it. Does anyone know how to start solving it?
Question : Let $X_1$ and $X_2$ be independent exponentially distributed random variables with parameters $\lambda_1$ and $\lambda_2$ so that $Pr\{X_i>t\}=e^{-\lambda_it}$ for $t\ge0$.
Let
- $N=1$ if $X_1<X_2$, and
- $N=2$ if $X_2\le X_1$,
- $U=\min\{X_1, X_2\}=X_N$,
- $V=\max\{X_1,X_2\}$ and
- $W=V-U=|X_1-X_2|$.
Show that
a. $Pr\{N=1\}=\frac{\lambda_1}{\lambda_1+\lambda_2}$ and $Pr\{N=2\}=\frac{\lambda_2}{\lambda_1+\lambda_2}$
b. $Pr\{U>t\}=e^{-(\lambda_1+\lambda_2)t}$ for $t\ge0$
c. $N$ and $U$ are independent random variables
d. $Pr\{W>t|N=1\}=e^{-\lambda_2t}$ and $Pr\{W>t|N=2\}=e^{-\lambda_1t}$ for $t\ge0$
e. $U$ and $W=V-U$ are independent random variables.
Best Answer
For a), use the law of total probability: $$ {\rm P}(X_1 < X_2 ) = \int_0^\infty {{\rm P}(X_1 < X_2 |X_2 = t)f_{X_2 } (t)\,{\rm d}t} , $$ where $f_{X_2}$ is the PDF of $X_2$.
For b), notice that $\min \{ X_1 ,X_2 \} > t$ if and only if $X_1 > t$ and $X_2 > t$ (and use the fact that $X_1$ and $X_2$ are independent).
For c), calculate ${\rm P}(\min \{X_1,X_2 \}>t , X_1 > X_2 )$ using the law of total probability, conditioning on $X_2$. You should easily find that $$ {\rm P}(\min \{X_1,X_2 \}>t , X_1 > X_2 ) = {\rm P}(\min \{X_1,X_2 \}>t ){\rm P}(X_1 > X_2 ) = \frac{{\lambda _2 }}{{\lambda _1 + \lambda _2 }}e^{ - (\lambda _1 + \lambda _2 )t}. $$
For d), note that $$ {\rm P}(|X_1 - X_2 | > t|N = 1) = \frac{{{\rm P}(X_2 - X_1 > t,X_1 < X_2 )}}{{{\rm P}(X_1 < X_2 )}} = \frac{{{\rm P}(X_2 > X_1 + t)}}{{{\rm P}(X_1 < X_2 )}}, $$ and you should easily show using the law of total probability, conditioning on $X_1$, that $$ {P(X_2 > X_1 + t)} = \frac{{\lambda _1 }}{{\lambda _1 + \lambda _2 }}e^{ - \lambda _2 t}. $$ Note: The calculation for ${\rm P}(|X_1 - X_2 | > t|N = 2)$ is completely analogous.
NOTE: Since question e) is not so easy, I give more than hints. However, try solving a significant part of it by yourself.
For e), it is straightforward to show, using that $N$ and $U$ are independent, that $$ {\rm P}(W > t | U=u) = {\rm P}(W > t | N=1, U=u){\rm P}(N=1) + {\rm P}(W > t | N=2, U=u){\rm P}(N=2). $$ For this purpose, you may replace $U=u$ by $U \in [u,u+{\rm d}u]$, where ${\rm d}u \to 0$, in order to condition on events with positive probability. Now, given $U=u$ and $N=1$, we have that $X_1 = u$ and that $X_2 - X_1$, by standard property of the exponential distribution, is exponential$(\lambda_2)$. Analogously, given $U=u$ and $N=2$, we have that $X_2 = u$ and that $X_1 - X_2$ is exponential$(\lambda_1)$. From this you should find that $$ {\rm P}(W > t | U=u) = \frac{{\lambda _1 }}{{\lambda _1 + \lambda _2 }}e^{ - \lambda _2 t} + \frac{{\lambda _2 }}{{\lambda _1 + \lambda _2 }}e^{ - \lambda _1 t} . $$ Now we are done by $$ {\rm P}(W > t ) = {\rm P}(W > t ,N = 1) + {\rm P}(W > t ,N = 2), $$ as it gives us, by virtue of a) and d), $$ {\rm P}(W > t) = \frac{{\lambda _1 }}{{\lambda _1 + \lambda _2 }}e^{ - \lambda _2 t} + \frac{{\lambda _2 }}{{\lambda _1 + \lambda _2 }}e^{ - \lambda _1 t} . $$