Probability conditioning on $X= Y$

Question

Given a random variable $X\sim \text{Exp}(1)$, I would like to condition on the event that $$``\text{$X$ behaves like an $\text{Exp}(1/2)$ random variable}".$$ Is there a way to define this event and state this as a proper conditional probability?

I know the Radon-Nikodym derivative can be used to describe this, but that would require us to change the probability measure.

Is there a way to do this (maybe using the RN derivative?) as a conditional probability?

Edit: I changed the $\text{Exp}(2)$ to $\text{Exp}(1/2)$ from the original question, which is more related to what I am thinking about. I didn't realize that there is a big difference previously.

Update: I think there is a partial answer from this post.

Due to a result by Diaconis and Zabell (Theorem 2.1 and the remark after):

If $Q\ll P$ and their RN-derivative $\frac{dQ}{dP}$ is bounded, then there exists a probability space such that $P$ is a marginal measure and $Q$ is a conditional measure.

I wonder if there are more recent results that handle the case when the RN-derivative is unbounded?

In the same paper, there is also the interesting remark

For example, no geometric distribution can be obtained from a Poisson distribution by conditioning, but any Poisson distribution can be obtained from any geometric.

But I do not quite see the reasoning.

Answer 1

Let $X_0 \sim \mathrm{Exp}(2)$, and $X_1$ be an independent random variable with density

$$f_{X_1}(x) = 2(e^{-x} - e^{-2x}), \qquad x \in [0, \infty).$$

Also let $Y \sim \mathrm{Bernoulli}(\tfrac12)$, independently, and set $Z=X_Y$. You can check that $Z$ has density

$$f_Z(x) = \tfrac12(f_{X_0}(x) + f_{X_1}(x)) = e^{-x},$$

so $Z \sim \mathrm{Exp}(1)$, and $Z \mid \{Y=0\} \sim \mathrm{Exp}(2)$.

(More generally, you could apply rejection sampling.)

Answer 2

I do not know whether you had something like this in mind, but you can do the following:

Introduce another random variable $Y\sim \text{Exp}(1)$ independent from $X$. Then by symmetry since $\Bbb P (X=Y)=0$: $$\Bbb P (X<Y) =\frac 1 2.$$ Also by symmetry $$\Bbb P ( \min (X,Y) > t, X<Y) = \Bbb P ( \min (X,Y) > t, X>Y)$$ Consequently, $$\Bbb P ( \min (X,Y) > t, X<Y) = \frac 1 2 \Bbb P ( \min (X,Y) > t)$$. This yields $$\Bbb P ( X > t\vert X<Y) = \Bbb P ( \min (X,Y) > t\vert X<Y) \\ = \frac{\Bbb P ( \min (X,Y) > t, X<Y)}{\Bbb P ( X<Y)} = \Bbb P ( \min (X,Y) > t)$$ Since the minimum of two independent exponential random variables is again exponentially distributed with rate equal to the sum of the previous rates you obtain $$\Bbb P ( X > t\vert X<Y) = \exp (-2t).$$ This means $X$ conditioned on $X<Y$ has distribution $\text{Exp}(2)$.