PDF of sum of exponentials conditioned on equality.

Question

I am trying to solve the following exam practice problem:

Let $X_1, X_2$ be independent exponential random variables with parameter $1$.

1. Find the conditional PDF of $X_1+X_2$ given that $\frac{X_1}{X_2}=1.$
2. Find the conditional PDF of $X_1+X_2$ given that $X_1-X_2=0.$
3. The events $\frac{X_1}{X_2}=1$ and $X_1-X_2=0$ are the same. Does this mean that conditioning on either of these two events should give the same answer?

Here is my approach:

For part 1, I computed the joint PDF of $Y= X_1+X_2$ and $Z=\frac{X_1}{X_2}$ and my result was $$f_{Y,Z}(y,z) = \frac{ye^{-y}}{(z+1)^2}\mathbf{1}_{\{y\geq 0, z \geq 0\}}.$$ This shows that $Y$ and $Z$ are independent. So, conditioned on $Z=1$, we get that $$f_{Y|Z}(y|1) = ye^{-y}\mathbf{1}_{\{y\geq 0\}}.$$

For part 2, I computed the joint PDF of $U=X_1+X_2$ and $V=X_1-X_2$ and got $$f_{U,V}(u,v) = \frac{1}{2}e^{-u}\mathbf{1}_{\{u\geq |v|\}}.$$ I then used Bayes' theorem to get that $$f_{U|V}(u|0) = e^{-u}\mathbf{1}_{\{u\geq 0\}}.$$

My question regards part 3. Firt of all, part 3 states that the events $X_1-X_2=0$ and $\frac{X_1}{X_2}=1$ are the same. But, I disagree because the event $X_1-X_2=0$ contains the measure-zero event $X_1=X_2=0$, which is not contained in the event $\frac{X_1}{X_2}=1$. Is this not the case? Second of all, if these two events were indeed the same, then why would conditioning on them produce a different result?

Answer 1

I tend to feel that conditioning on $X_1=X_2$ is not really consistent with the assumption that $X_1$ and $X_2$ are independent, so seeing things like the Borel–Kolmogorov paradox should not be a surprise.

You might say that conditioning on an event of probability $0$ is meaningless in itself as it leads to $\frac00$, but can only be interpreted as the limit of conditioning on an event of positive probability. Here

• the first is the limit of the distribution of $X_1+X_2$ conditioned on $(1-\epsilon_a)X_2 \le X_1 \le (1+\epsilon_b)X_2$ and
• the second is the limit of the distribution of $X_1+X_2$ conditioned on $X_2-\epsilon_c \le X_1 \le X_2 +\epsilon_d$

as $\epsilon_a,\epsilon_b,\epsilon_c,\epsilon_d \to 0_+$, and there is no reason to expect the two limiting distributions to be the same.

The first constraint tends to cut out small values of pairs of $X_1$ and $X_2$ which have large relative differences but small absolute differences and to leave in large values of such pairs which have small relative differences but large absolute differences, while the second constraint does the opposite, so it is not particularly a surprise that for small $\epsilon$s the first has a mean for $X_1+X_2$ near $2$ while the second has a mean near $1$, each consistent with your earlier answers.

Using the following R code to simulate this with $\epsilon=0.1$

set.seed(2019)
cases <- 10^6
epsilon <- 0.1
X1 <- rexp(cases, 1)
X2 <- rexp(cases, 1)
Y <- X1 + X2
YA <- Y[(1-epsilon) * X2 <= X1 & X1 <= (1+epsilon) * X2]
YB <- Y[X2 - epsilon     <= X1 & X1 <= X2 + epsilon]
c(length(YA), length(YB)) # numbers of observations 
# 50187 95015
c(mean(YA), mean(YB)) # means of observations
# 1.995142 1.048147
plot.ecdf(YA, col="red")
plot.ecdf(YB, col="blue", add=TRUE)

you see the different simulated means are near $2$ and $1$ and the different empirical cumulative distribution functions are shown below in red and blue, broadly consistent with your answers.

The next chart of some of the simulated pairs helps illustrate how the red constraint tends to select more larger values and fewer smaller values than the blue constraint

To me it is curious that your first answer is a $\text{Gamma}(2,1)$ distribution while your second is an exponential distribution with mean $1$, while I would have had an intuitive attraction to an answer which was an exponential distribution with mean $2$, i.e. the distribution of $2X_1$