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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

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2 Answers 2

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) $a=0.47$, $b=0.92$, $x=0.81$. $\hat F_R (x) = 0.2926322$, $F_R (x) = 0.2926885...$.

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Confirmed using simulations. Examples later on... –  Shai Covo Jul 11 '11 at 15:43
    
Perhaps it should be noted that the distribution function has a discontinuity at $x=1$, so that the derivative of the above expression does not integrate to $1$. –  Jason Swanson Jul 11 '11 at 15:44
    
@user11867: Good observation. I'll add details on this later on. Anyway, the expression for the distribution function should be correct. –  Shai Covo Jul 11 '11 at 15:49
    
Details added... –  Shai Covo Jul 11 '11 at 16:42
    
Also added examples confirming the expression for the distribution function. –  Shai Covo Jul 11 '11 at 19:33

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 $R \leq 1$) $$ F_R (1) = 1. $$

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for this question pdf of min (x/y , 1) = pdf of min [(x,y)]/y or not?? Again that x and y are exponential random variable –  Wanakorn Swasdio Jul 13 '11 at 11:52
    
$\min(X/Y,1)=\min(X,Y)/Y$, in particular the corresponding pdf's are equal. This should lead to a simpler derivation of the distribution function. I'll give the details later on. –  Shai Covo Jul 13 '11 at 13:35
    
The details are given in the Edit... –  Shai Covo Jul 13 '11 at 16:05

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