# What is the probability that $X \geq k\cdot Y$, if X and Y are independent and follow the exponential distribution?

$X$ and $Y$ are independent variables that follow the exponential distribution with the same parameter $\beta$. What is the probability $$P(X \geq k\cdot Y)$$ if $k>0$ ?

I don't really know where to start.

• Set up and evaluate a double integral. – André Nicolas Jun 9 '16 at 13:38

$X$ and $Y$ follow an exponential distribution. Therefore, the density function is $f(t)= \beta \exp{(-\beta t)}$. First, we condition by $Y=k$ and integrates over the $k$.
$$P(X\geqslant kY) = \int_0^\infty \beta \exp{(-\beta t)} P(X\geqslant kt| Y=t)dt$$
Since $X$ and $Y$ are independent, $P(X\geqslant kt | Y=t)=P(X\geqslant kt)=\exp(-\beta \times kt)$. Plugging it in the previous equality: $$P(X\geqslant kY) = \int_0^\infty \beta \exp{(-\beta (k+1)t)}dt$$ Which by integration gives: $$\frac{\beta}{\beta (k+1)}=\frac{1}{k+1}$$
Required probability $$\iint_{x\ge ky\ge0}p_{X,Y}(x,y)dA=\int_{0}^\infty\int_{ky}^\infty p_{X,Y}(x,y)dxdy\\ =\int_{0}^\infty\int_{ky}^\infty p_X(x)p_Y(y)dxdy=\beta^2\int_{0}^\infty\int_{ky}^\infty e^{-\beta(x+y)}dxdy\\ =\beta\int_0^{\infty}-e^{-\beta(x+y)}\Big{|}_{ky}^\infty dy=\beta\int_0^{\infty}e^{-\beta(ky+y)} dy={1\over k+1}$$
Hint: For fixed $k>0$, $$\Pr(X\geq kY)=\int_0^{\infty}\Pr(X\geq ky)f_Y(y)\;dy$$