Proof independence of $Z=min(X_1,X_2,....X_n)$ and $J:X_{j=J}=Z$, where each $X$ is exponential distributed

Question

Suppose $X_{i}$ for $i=1,2,..,n$ are independent random variables. Let their distribution be exponential with parameter $\lambda_{i}$ respectively. and $Z=min(X_{1},X_{2},....X_{n})$ and J= j where j is the index of random variable which is equal to Z . How can we prove that Z and J are independent ?

My try at the solution

TO prove $J$ and $Z$ are independent it will suffice to prove $P(J,Z)=P(J)P(Z)$

I start by trying to find the joint probability of $J=k$ and $Z\geq t$ , So I have $$P(J=k, Z\geq t)=P(J=k|Z\geq t)*P(Z\geq t)$$ But Z is the minimum of all the random variables so every random variable $X_{i}$ is greater than or equal to $t$ which gives us $$P(Z\geq t)=P(X_{1}\geq t, X_{2}\geq t,......,X_{n}\geq t)$$

And since $Z_{i}$ are independent random variable we have

$$P(Z\geq t)=P(X_{1}\geq t)P(X_{2}\geq t)...P(X_{n}\geq t)$$

I can't seem to figure out how should I proceed next. I will have to particularly play with the conditional term and see if I could make it just $P(J=k)$ or maybe I should plug in the probabilities of exponential random variables in the last equation. Can anybody help, give me any hints ?

Answer 1

joint probability $Z=z$ and $J=j$

$$\begin{align} \\P(Z=z , J=j) &= f_j(Z) \prod_{i \neq j} 1-F_i(z)\\& = \lambda_je^{-\lambda_j z} \prod_{i \neq j} 1 - (1-e^{-\lambda_i z}) \\&= \lambda_j \prod_{i} e^{-\lambda_i z}\\&=\lambda_j e^{- \left( \sum_{i}\lambda_i \right) z} \end{align}$$

joint probability $X_j=x$ and $J=j$ works the same

$$\begin{align} \\P(X_j=x , J=j) &= \lambda_j e^{- \left( \sum_{i}\lambda_i \right) x}\\ &\phantom{=\lambda_je^{-\lambda_j z} \prod_{i \neq j} 1 - (1-e^{-\lambda_i z})} \end{align}$$

we can find the probability for $J=j$ by integrating the previous result

$$\begin{align} P(J=j) &=\int_0^\infty P(X_j=x,J=j) dx \\ &= \frac{\lambda_j}{-\sum_{i}\lambda_i } e^{- \left( \sum_{i}\lambda_i \right) x} \Big|_0^\infty\ \\ &= \frac{\lambda_j}{\sum_{i}\lambda_i } \end{align} $$

Thus $P(Z=z \vert J=j) = \frac{P(Z=z , J=j)}{P(J=j)} $ is constant. Namely, we get:

$$P(Z=z \vert J=j) = \left( \sum_{i}\lambda_i \right) e^{- \left( \sum_{i}\lambda_i \right) z} $$

and the conditional distribution is independent from the index $j$

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checking by computation:

# lamda_i
lambda = c(1,2,3)

# sampling
n <- 100000
sample <- sapply(lambda,function(l) rexp(n,l))

# descriptive statistics
min <- apply(sample,1,min)
id  <- apply(sample,1,function(x) which.min(x))

# observations
j=1
plot(hist(min[which(id==j)],breaks=seq(0,max(min),0.05)),col=2)

# plotting model
x <- seq(0,4,0.05)
lines(x,0.05*length(which(id==j))*sum(lambda)*exp(-sum(lambda)*x))

which seems to work well