What is the CDF of $X\sim\text{EXP}(\frac1y)$, where $Y\sim\text{EXP}(1)$?

Question

X, Y are two independent random variables such that for a given $Y=y$, $X \sim \text{EXP}(\theta = \frac{1}{y})$ and $Y \sim \text{EXP} (\theta = 1)$.

I thought about using the law of total probability and then integrating from $-\infty$ to $x$, but I cannot seem to solve it:

$$\text{P} (A) = \text{E} (\text{P} (B|A) )$$

If $A = X$ and $B = Y$, then

$$\text{P} (X) = \text{E} (\text{P} (Y|X))$$

Because $Y$ does not depend on $X$, $\text{P} (Y|X) = \text{P} (Y)$. But then we get:

$$\text{P} (X) = \text{E} (\text{P}(Y))$$

I don't really know what to do next.

Answer 1

Your instinct to use the total law of probability is correct. The CDF is $\mathbb{P}(X \leq x)$, so try to use the law of total probability to calculate it. Hover below to see the calculation

$$\mathbb{P}(X \leq x) = \mathbb{E}\left[\mathbb{P}(X \leq x \, | \, Y) \right] = \mathbb{E}\left[1 - e^{-Y x} \right] = 1 - \frac{1}{1 + x}$$