# Probability of ($x\le 2Y$)

$X$ and $Y$ are independent and their probability density functions are $$f_X(t)=f_Y(t)=\left\{\begin{array}{l} e^{-t},\:\text{if t \geq 0;} \\ 0,\:\text{otherwise.}\end{array}\right.$$

$P(X \leq 2Y)$=? (The probability of $X \leq 2Y$)

Since the two variables are independent, their joint density function is just the product of their marginal densities, i.e. $$f(x,y) = e^{-x}e^{-y} \;\ x,y > 0$$ In order to find $P(X\leq2Y)$, we need to solve $$\int_{0}^{\infty}\int_{0}^{2Y}f(x,y)\;\ dx \;\ dy$$ since $X$ must be less than or equal to $2Y$, and we are considering all possible values of $Y$. Evaluating this we have: $$\int_{0}^{\infty} e^{-y} \;\ \{\int_{0}^{2Y}e^{-x}dx\} \;\ dy \; = \int_{0}^{\infty} e^{-y} (1 - e^{-2y}) \;\ dy =$$
$$\int_{0}^{\infty}e^{-y}dy - \int_{0}^{\infty}e^{-3y}dy = 1 - \frac{1}{3} = \frac{2}{3}$$ Note that the last integral, $\int_{0}^{\infty}e^{-3y}dy$, can be thought of as the PDF of an exponential random variable with mean $1/3$, except since the scaling factor of 3 was omitted, the integral evaluates to only $\frac{1}{3}$ of the area of a proper density function, 1. When you encounter a problem like this concerning two random variables and the probability of some inequality of them, i.e. $X \leq 2Y$, $X > Y + 4$, etc... you should expect to be solving a double integral. The only tricky part is setting up the bounds of the integral. It is often helpful to draw a graph relating the two variables so you can visually identify the region of interest and simply take the boundaries of that.
• A quick check on your calculations (in R) confirms that this answer: pf(2,2,2) yields [1] 0.6666667. – Glen_b -Reinstate Monica May 3 '14 at 23:24