Conditional distribution of $X$ given $Y$

Question

I'm practicing on conditional distribution. This is my exercise

Suppose that the lifespans of two light bulbs are independent and identically distributed exponential RVs. Let $X$ and $Y$ be the times at which the first and second bulbs burn out, respectively ($X \leq Y$). What is the conditional distribution of $X$ given $Y$.

Can I do something like this: Since $X,Y$ are independent, then $f_{X,Y}(x,y) = f_X(x).f_Y(y)$. Then $f_{X}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(y)} = f_X(x)$. I think that I must be so wrong about this. Can somebody please help me out.

Answer 1

Let the lifespans of the bulbs be $A$ and $B$. These are iid exponential RVs. Then $X = \min\{A, B\}$ and $Y = \max\{A, B\}$. When I put it like that, maybe it's easier to see why $X$ and $Y$ are not independent, which is where your attempted solution breaks.

To actually solve the problem, we can try to understand the joint distribution of $X$ and $Y$. Suppose $A, B \sim Exp(\lambda)$. For $s \le t$ the joint CDF will be $$\begin{align} F_{X,Y}(s,t) &= P[X \le s, Y \le t] \\ &= P[A \le t \text{ and } B \le t] - P[s < A \le t \text{ and } s < B \le t] \\ &= F_A(t) F_B(t) - [F_A(t)-F_A(s)][F_B(t)-F_B(s)] \\ &= \left( 1 - e^{-\lambda t}\right)^2 - \left( e^{-\lambda s} - e^{-\lambda t}\right)^2 \end{align}$$

Then for the joint pdf we get $$\begin{align} f_{X,Y}(s,t) &= \frac{d^2}{dsdt} F_{X,Y}(s,t) \\ &= 2 \lambda^2 e^{-(s+t)\lambda} \end{align}$$

Also, the CDF for $Y$ is $F_Y(t) = F_{X,Y}(t, t)$, so the pdf is $$\begin{align} f_{Y}(t) &= \frac{d}{dt} \left( 1 - 2 e^{-\lambda t} + e^{-2 \lambda t} \right) \\ &= 2 \lambda \left( e^{-\lambda t} - e^{-2 \lambda t} \right) \end{align}$$

Now we're finally ready to plug into the formula for conditional pdf: $$\begin{align} f_{X|Y}(x|y) &= \frac{f_{X,Y}(x,y)}{f_{Y}(y)} \\ &= \frac{2 \lambda^2 e^{-(x+y)\lambda}} {2 \lambda \left( e^{-\lambda y} - e^{-2 \lambda y} \right)} \\ &= \boxed{\frac{\lambda e^{\lambda(t-s)}}{e^{\lambda t} - 1}} \end{align}$$

Note that the final formula, like all the other cdfs and pdfs computed in this solution, is only valid on the region $0 \le s \le t$. Outside of that region the pdf is 0.