# Independent random variables and CDF

Let $$X,Y,Z$$ are independent random variables, such that $$X,Y$$ have exponential distribution with the same parameter $$\alpha$$ and $$Z$$ has Bernoulli distribution $$P(Z=0)=p$$ and $$P(Z=1)=1-p$$. Find CDF $$W=\frac{X}{X+YZ}$$.

My solution:

$$P(W \le t)=P(W \le t \wedge Z=0) + P(W \le t \wedge Z=1)= P(1 \le t \wedge Z=0)+P(\frac{X}{X+Y} \le t \wedge Z=1)=pP(1 \le t)+(1-p)P(\frac{X}{X+Y} \le t)$$.

For $$t<0$$ we have $$0$$.

For $$0 \le t <1$$ we have $$1-p$$

For $$t \ge 1$$ we have $$p$$.

Did I do it correctly?

Suppose X and Y are independently distributed exponential random variables with parameter $$\alpha$$. This means $$f_{X}(x) = \alpha e^{-\alpha x}$$ What is the distribution of $$U=\frac{X}{X+Y}$$ Here is how one might proceed. Firstly note that U must be between 0 and 1. For t(0,1),

$$P(\frac{X}{X+Y}\le t) = P(\frac{X+Y}{X}\ge \frac{1}{t})$$

$$=P(Y\ge x(\frac{1}{t}-1))$$ $$=\int_{0}^{\infty} f_X(x)Pr\left(Y\ge x(\frac{1}{t}-1)\right)$$

$$\int_{0}^{\infty} \alpha e^{-\alpha x} . \left(1-e^{-\alpha. x(\frac{1}{t}-1))}\right) dx$$

$$\int_{0}^{\infty}\left(\alpha e^{-\alpha x} -\alpha e^{\frac{-\alpha x}{t}}\right) = 1-t$$

Now your solution is complete as @drhab said. Goodluck

• Errr... exponential random variables with density $ae^{-ax}$ for $x \geq 0$ have mean $\frac 1a$, not mean $a$. – Dilip Sarwate Jan 19 at 16:27
• I will change it to parameter $\alpha$ and keep the expression the same rather change it by reversing it all through as it does not change the result. Thanks for noting that – Satish Ramanathan Jan 19 at 16:29

For $$0\leq t<1$$ we do not have $$1-p$$ but $$(1-p)P\left(\frac{X}{X+Y}\leq t\right)$$.
For $$t\geq1$$ we do not have $$p$$ but $$p\cdot1+(1-p)\cdot1=1$$.
To be found is yet $$P\left(\frac{X}{X+Y}\leq t\right)$$ for $$0\leq t<1$$ so your answer is also not complete.