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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

2 Answers 2

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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...2011-07-11
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    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 $1$.2011-07-11
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    @user11867: Good observation. I'll add details on this later on. Anyway, the expression for the distribution function should be correct.2011-07-11
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    Details added...2011-07-11
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    Also added examples confirming the expression for the distribution function.2011-07-11
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    Thank you so much for your derivation ^^2011-07-12
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    I will combine this pdf with my total equation and integral it again to see the performance of my system If it has not any problem you can see it in my wireless communication journal2011-07-12
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    @Wanakorn: Glad to help.2011-07-12
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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 variable2011-07-13
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    $\min(X/Y,1)=\min(X,Y)/Y$, 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.2011-07-13
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    The details are given in the Edit...2011-07-13