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

  • $\begingroup$ Errr... exponential random variables with density $ae^{-ax}$ for $x \geq 0$ have mean $\frac 1a$, not mean $a$. $\endgroup$ – Dilip Sarwate Jan 19 at 16:27
  • $\begingroup$ 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 $\endgroup$ – Satish Ramanathan Jan 19 at 16:29

Your start is okay but:

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.


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