Find $P(Z \ge \frac{1}{3} )$

Question

Given two independent random variables, $X \sim Exp(2)$, $ Y \sim Exp(1) $. Define $Z=\frac{X}{X+Y}$, I need to find $P(Z \ge \frac{1}{3} )$.

I expanded this:

$$ P(Z \ge \frac{1}{3} ) = P(\frac{X}{X+Y} \ge \frac{1}{3})=P(X \ge \frac{1}{3}X + \frac{1}{3}Y) = P(\frac{2}{3}X \ge \frac{1}{3}Y) = P(2X \ge Y)$$

And I don't know how to continue.

Answer 1

$P(2X \geq Y)=\int_0^{\infty} \int_0^{2x}e^{-y}dy 2e^{-2x} dx=\frac 1 2$.

Answer 2

You want to calculate $P(2X - Y \geq 0) = P(W + Z \geq 0) = P(S \geq 0) = 1 - P(S < 0)$ where $W \sim \mathrm{Exp}(1)$ and $f_Z(z) = 0$ when $ z \geq 0$ and $f_Z(z) = e^{z}$ when $z < 0$. $f_Z(z)$ is pdf (Probability Density Function) of Z. The distribution of $S$ can be calculated using convolution $$f_{S}(s)=\int_{-\infty}^{\infty} f_{Z}(z)f_{W}(s-z)dz = \int_{z<0\\ z<s} e^{z}e^{-(s-z)}dz$$ If $s < 0$ then $p_{S}(s)=\int_{-\infty}^{s}e^{z}e^{-(s-z)}dz=e^{-s}\int_{-\infty}^{s}e^{2z}dz=\frac{1}{2}e^{s}$. Using this result $$1 - P(S < 0) = 1 - \int_{-\infty}^{0} \frac{1}{2} e^{s}ds=\frac{1}{2}$$

To show that $W \sim Exp(1)$ calculate cdf (Cumulative Distribution Function) of $W$ $$P(W \leq w) = P(X \leq w/2) = 1 - e^{-2\frac{w}{2}} = 1 - e^{-w}$$

pdf of Z can also be established from cdf $P(Z \leq z) = P(-Y \leq z) = P(Y \leq -z)$.

Answer 3

We can use the following property:

If $X\sim \mathop{Exp}(\lambda)$ and $Y\sim \mathop{Exp}(\mu)$ are independent, then $P(X<Y)=\frac{\lambda}{\lambda+\mu}$.

Since $X\sim \mathop{Exp}(2)$, we have $2X\sim \mathop{Exp}(1)$ (can you prove this?). Now, applying above property, we obtain: $$P(2X\geq Y)=1-P(2X<Y)=1-\frac{1}{1+1}=\frac{1}{2}.$$