Let $X$ and $Y$ be i.i.d. Expo($\lambda$). Find the conditional distribution of $X$ given $X < Y$.

Question

(a) by using calculus to find the conditional PDF.

My Solution:

$$P(X \leq t \mid X < Y) = \dfrac{P(X \leq t, X < Y)}{P(X < Y)}\\ = \frac{P(X \leq t, X < Y \mid Y \leq t)P(Y \leq t) + P(X\leq t, X < Y\mid Y > t)P(Y>t)}{P(X < Y)}\\ = \frac{P(X < Y)P(Y\leq t) + P(X\leq t)P(Y > t)}{P(X < Y)} = 1 - e^{-\lambda t} + 2(1 - e^{-\lambda t})e^{-\lambda t}\\ =1 - e^{-\lambda t}-2e^{-2\lambda t}$$

I was going to take the derivative with respect to $t$ to find the conditional PDF, but realized my CDF is incorrect since the text claims $$ P(X \leq t\mid X < Y) = P(\min(X, Y) \leq t) $$ Which means $$ P(X \leq t\mid X < Y) = P(\min(X, Y) \leq t) = 1- e^{2\lambda t} $$

Any help would be greatly appreciated.

Answer 1

Rather, try this:

$$\begin{align}\mathsf P(X\leqslant t\mid X\lt Y) &= \dfrac{\mathsf P(X\lt Y, X\leqslant t)}{\mathsf P(X\lt Y)}\\[1ex] &=\dfrac{\mathsf P(X\lt Y, Y\leqslant t, X\leqslant t)+\mathsf P(X\lt Y, X\leqslant t, Y>t)}{\mathsf P(X\lt Y)}\\[1ex] &=\dfrac{\mathsf P(X\lt Y\mid X\leqslant t, Y\leqslant t)\,\mathsf P(X\leqslant t,Y\leqslant t)+\mathsf P(X\leqslant t, Y>t)}{\mathsf P(X\lt Y)}\\[1ex] &=\mathsf P(X\leqslant t)\,\mathsf P(Y\leqslant t)+2\,\mathsf P(X\leqslant t)\,\mathsf P(Y>t)\\ & ~~\vdots\end{align}$$

Answer 2

I don´t know why the min function has been used. But you can just calculate the corresponding integrals

$$P(X<Y\cap X\leq t)=\int_0^t \left[ \int_x^{\infty} \lambda\cdot e^{-\lambda y} \, dy \right] \lambda\cdot e^{-\lambda x} \, dx$$

$$=\int_0^t \left[ - e^{-\lambda y} \, dy \right]_x^{\infty} \lambda\cdot e^{-\lambda x} \, dx=\int_0^t e^{-\lambda x} \cdot \lambda\cdot e^{-\lambda x} \, dx$$

$$\lambda\cdot\int_0^t e^{-2\lambda x} \, dx=\lambda\cdot\left[ -\frac1{2\lambda}\cdot e^{-2\lambda x}\right]_0^{t}$$

$$=\frac12\cdot \left[ - e^{-2\lambda x}\right]_0^{t}=\frac12\cdot \left( 1-e^{-2\lambda t}\right)$$

The denominator can be calculated by using integrals as well

$$P(X<Y)=\int_0^{\infty} \left[ \int_0^{y} \lambda\cdot e^{-\lambda x} \, dx \right] \lambda\cdot e^{-\lambda y} \, dy$$

Or just using the symmetry. Since $X$ and $Y$ are equal distributed $P(X<Y)=\frac12$