How to find the pdf of [min(RV.1,RV.2)]/RV.2 RV.1=exponential random variable 1
RV.2=exponential random variable 2
both of them are independent
pdf of [min(RV.1,RV.2)]/RV.2=?????
pdf of min(RV.1,RV.2)=(a+b) exp (a+b) 
where a and b is the rate parameter of that exponential
Another choice 
pdf of min(RV,1)=???
I need this pdf to slove the problem in wireless communication CMRC
Thank you very much
 A: Suppose that $X$ and $Y$ are independent exponential rv's, with densities $ae^{-au}$ and $be^{-bu}$, $u > 0$, respectively.
For any $0 < x < 1$ and $s > 0$,
$$
{\rm P}(\min (X,s) \le xs) = {\rm P}(X \le xs) = 1 - e^{ - axs} .
$$
For any $0 < x < 1$, by the law of total probability (conditioning on $Y$)
$$
{\rm P}\bigg(\frac{{\min (X,Y)}}{Y} \le x \bigg) = \int_0^\infty  {{\rm P}\bigg(\frac{{\min (X,s)}}{s} \le x \bigg)be^{ - bs} \,ds}  = \int_0^\infty  {{\rm P}(\min (X,s) \le xs)be^{ - bs} \,ds} ,
$$
hence
$$
{\rm P}\bigg(\frac{{\min (X,Y)}}{Y} \le x \bigg) =  \int_0^\infty  {(1 - e^{ - axs} )be^{ - bs} du}  = 1 - b\int_0^\infty  {e^{ - (ax + b)s} ds}  = 1 - \frac{b}{{ax + b}}.
$$
The probability density function is obtained by differentiating the right-hand side.
EDIT (thanks to user11867's comment below): The ratio, call it $R$, has positive mass at $x=1$; hence, the density of $R$ on $(0,1)$ does not integrate to $1$. Specifically, the density $f_R$ of $R$ on $(0,1)$ is given by
$$
f_R (x) = \frac{d}{{dx}}\bigg(1 - \frac{b}{{ax + b}}\bigg) = \frac{{ab}}{{(ax + b)^2 }},\;\; 0 < x <1.
$$
It holds
$$
\int_0^1 {f_R (x)\,dx}  = \bigg(1 - \frac{b}{{ax + b}}\bigg) \bigg|_0^1  = 1 - \frac{b}{{a + b}} ,
$$
which is less than $1$. This implies that ${\rm P}(R=1)=b/(a+b)$. Indeed, 
$$
{\rm P}(R=1)={\rm P}(\min (X,Y) = Y) = {\rm P}(Y \le X),
$$
and so, by the law of total probability (conditioning on $X$)
$$
{\rm P}(R=1) = \int_0^\infty  {P(Y \le u)ae^{ - au}\, du}  = \int_0^\infty  {(1 - e^{ - bu} )ae^{ - au} \,du}  = 1 - a\int_0^\infty  {e^{ - (a + b)u} \,du}  = 1 - \frac{a}{{a + b}} = \frac{b}{{a + b}}.
$$
To summarize: The ratio $R$ is a random variable supported on $[0,1]$. It has distribution function $F_R$ given by
$$
F_R (x) = 1 - \frac{b}{{ax + b}}, \;\; 0 \leq x < 1,
$$
and
$$
F_R (1) = 1.
$$
Thus $F_R$ has jump discontinuity at $x=1$:
$$
F_R (1) - \mathop {\lim }\limits_{x \to 1^ -  } F_R (x) = 1 - \mathop {\lim }\limits_{x \to 1^ -  } \bigg(1 - \frac{b}{{ax + b}}\bigg) = \frac{b}{{a + b}},
$$ 
which is the probability ${\rm P}(R=1)$. In particular, the density function of $R$ exists only for $x < 1$; it is given by 
$$
f_R (x) = \frac{{ab}}{{(ax + b)^2 }},\;\; 0 < x < 1.
$$
EDIT 2: The result 
$$
F_R (x) = 1 - \frac{b}{{ax + b}}, \;\; 0 \leq x < 1,
$$
can be easily confirmed using Monte Carlo simulations, using the fact that an exponential random variable with density function $\lambda e^{-\lambda x}$, $x > 0$, can be generated as $-\ln(U)/\lambda$, where $U$ is a uniform$(0,1)$ random variable. (Indeed, it is straightforward to check that ${\rm P}(-\ln(U)/\lambda \leq x) = 1-e^{-\lambda x}$, for any $x > 0$.) Let $\hat F_R (x)$, for $0 < x < 1$ fixed, denote the Monte Carlo approximation for $F_R (x)$. The following results were obtained (using $N=10^7$ repetitions for each approximation; better approximations can be obtained by increasing $N$):
1) $a=0.8$, $b=1.34$, $x=0.53$. $\hat F_R (x) = 0.2401641$, $F_R (x) = 0.2403628...$;
2) $a=3.2$, $b=0.85$, $x=0.28$. $\hat F_R (x) = 0.5135828$, $F_R (x) = 0.5131729...$;
3) $a=2.4$, $b=5.18$, $x=0.21$. $\hat F_R (x) = 0.0885315$, $F_R (x) = 0.0886699...$;
4) $a=0.47$, $b=0.92$, $x=0.81$. $\hat F_R (x) = 0.2926322$, $F_R (x) = 0.2926885...$.
A: Concerning the second question, let 
$$
M = \min (X,1),
$$
where $X$ is exponential with density function $ae^{-au}$, $u > 0$. Then $M$ is a random variable supported on $[0,1]$.
For any $0 < x < 1$ fixed,
$$
{\rm P}(M \leq x) = {\rm P}(\min (X,1) \leq x) = {\rm P}(X \leq x) = 1 - e^{-ax}.
$$
On the other hand,
$$
{\rm P}(M=1) = {\rm P}(\min (X,1) = 1 ) = {\rm P}(X > 1) = e^{-a}.
$$
Let $F_M$ denote the distribution function of $M$. It is given by
$$
F_M (x) = 1 - e^{-ax}, \;\; 0 \leq x < 1,
$$
and
$$
F_M (1) = 1.
$$
Note that $M$ has a jump discontinuity at $x=1$:
$$
F_M (1) - \mathop {\lim }\limits_{x \to 1^ -  } F_M (x) = 1 - (1 - e^{ - a} ) = e^{-a},
$$
which is the probability ${\rm P}(M=1)$. In partcular, the density function of $M$ exists only for $x < 1$; it is given by
$$
f_M (x) = ae^{-ax}, \;\; 0 < x < 1.
$$
(Of course, it does not integrate to $1$.)
EDIT: Returning to the first question, the distribution function, $F_R$, of the ratio
$$
R = \frac{{\min (X,Y)}}{Y}
$$
can be derived simply as follows. The key observation (thanks to Wanakorn's comment below) is that
$$
\frac{{\min (X,Y)}}{Y} = \min \bigg(\frac{X}{Y},1 \bigg).
$$
Hence, for any $0 < x < 1$,
$$
{\rm P}(R \le x) = {\rm P}\bigg(\min \bigg(\frac{X}{Y},1 \bigg) \le x \bigg) = {\rm P}\bigg(\frac{X}{Y} \le x \bigg) = {\rm P}(X \le xY).
$$
Thus, by the law of total probability (conditioning on $Y$),
$$
{\rm P}(R \le x) = \int_0^\infty  {{\rm P}(X \le xu)be^{ - bu} \,du}  = \int_0^\infty  {(1 - e^{ - axu} )be^{ - bu} \,du}  = 1 - b\int_0^\infty  {e^{ - (ax + b)u} \,du}  = 1 - \frac{b}{{ax + b}}.
$$
Hence $F_R$ is given by
$$
F_R (x) = 1 - \frac{b}{{ax + b}}, \;\; 0 \leq x < 1,
$$
and (noting that $R \leq 1$)
$$
F_R (1) = 1.
$$
