pdf of product of two function of the Exponential variables
$begingroup$
I would like to find this probability: $Pr(Z<2^{2R})$ for $R>0$.
So, I try several ways and finally decide to find pdf of Z.
If $X$ and $Y$ are independent and exponentially distributed with parameter $lambda_1$ and $lambda_2$, respectively, and $a_i>0$ for $i=1,...,6$, which is the pdf of $Z$? Where $Z$ is given by
$$Z=frac{(a_1X+a_2)(a_3Y+a_5)}{(a_3X+a_4)(a_1Y+a_6)} $$
any idea?
First,considering $Z=W_{1}W_{2}$, I calculate pdf of $W1=frac{(a_1X+a_2)}{(a_3X+a_4)}=frac{W_{11}}{W_{12}}$ as:
$f_{W_{11}}(w_{11})=frac{1}{a_{1}}f_{X}(frac{w_{11}-a_{2}}{a_{1}})$ for $w_{11}>a_{2}$
$f_{W_{12}}(w_{12})=frac{1}{a_{3}}f_{X}(frac{w_{12}-a_{4}}{a_{3}})$ for $w_{12}>a_{4}$
$F_{W_{1}}(W_{1})=Pr(W1=frac{W_{11}}{W_{12}}<w_{1})=int_{a_{4}}^{infty}int_{a_{2}}^{ww_{2}} f_{w_{11}}(w_{11})f_{w_{12}}(w_{12}), dw_{11}dw_{12}=1-frac{{{e}^{{{lambda }_{1}}(frac{{{a}_{2}}}{{{a}_{1}}}+frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{lambda }_{1}}(frac{{{a}_{4}}}{{{a}_{3}}}+frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{(1+frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})}$
If we derive from the above, pdf of $W_{1}$ is achieved as:
$frac{a_{3}{{e}^{{{lambda }_{1}}(frac{{{a}_{2}}}{{{a}_{1}}}+frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{lambda }_{1}}(frac{{{a}_{4}}}{{{a}_{3}}}+frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{a_{1}(1+frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})^{2}}+frac{lambda_{1} a_{4}{{e}^{{{lambda }_{1}}(frac{{{a}_{2}}}{{{a}_{1}}}+frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{lambda }_{1}}(frac{{{a}_{4}}}{{{a}_{3}}}+frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{a_{1}(1+frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})}$
But, I have a problem: I know that integral of $f_{W_{11}}(W_{11})$ over whole interval equal to $1$. What are the integral bounds?(I need these bounds for calculating pdf of product of $W_{1}W_{2}$)
probability probability-distributions random-variables
asked Dec 9 '18 at 19:58
fatimaaafatimaaa
12
$endgroup$
add a comment |
$begingroup$
I would like to find this probability: $Pr(Z<2^{2R})$ for $R>0$.
So, I try several ways and finally decide to find pdf of Z.
If $X$ and $Y$ are independent and exponentially distributed with parameter $lambda_1$ and $lambda_2$, respectively, and $a_i>0$ for $i=1,...,6$, which is the pdf of $Z$? Where $Z$ is given by
$$Z=frac{(a_1X+a_2)(a_3Y+a_5)}{(a_3X+a_4)(a_1Y+a_6)} $$
any idea?
First,considering $Z=W_{1}W_{2}$, I calculate pdf of $W1=frac{(a_1X+a_2)}{(a_3X+a_4)}=frac{W_{11}}{W_{12}}$ as:
$f_{W_{11}}(w_{11})=frac{1}{a_{1}}f_{X}(frac{w_{11}-a_{2}}{a_{1}})$ for $w_{11}>a_{2}$
$f_{W_{12}}(w_{12})=frac{1}{a_{3}}f_{X}(frac{w_{12}-a_{4}}{a_{3}})$ for $w_{12}>a_{4}$
$F_{W_{1}}(W_{1})=Pr(W1=frac{W_{11}}{W_{12}}<w_{1})=int_{a_{4}}^{infty}int_{a_{2}}^{ww_{2}} f_{w_{11}}(w_{11})f_{w_{12}}(w_{12}), dw_{11}dw_{12}=1-frac{{{e}^{{{lambda }_{1}}(frac{{{a}_{2}}}{{{a}_{1}}}+frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{lambda }_{1}}(frac{{{a}_{4}}}{{{a}_{3}}}+frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{(1+frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})}$
If we derive from the above, pdf of $W_{1}$ is achieved as:
$frac{a_{3}{{e}^{{{lambda }_{1}}(frac{{{a}_{2}}}{{{a}_{1}}}+frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{lambda }_{1}}(frac{{{a}_{4}}}{{{a}_{3}}}+frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{a_{1}(1+frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})^{2}}+frac{lambda_{1} a_{4}{{e}^{{{lambda }_{1}}(frac{{{a}_{2}}}{{{a}_{1}}}+frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{lambda }_{1}}(frac{{{a}_{4}}}{{{a}_{3}}}+frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{a_{1}(1+frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})}$
But, I have a problem: I know that integral of $f_{W_{11}}(W_{11})$ over whole interval equal to $1$. What are the integral bounds?(I need these bounds for calculating pdf of product of $W_{1}W_{2}$)
probability probability-distributions random-variables
asked Dec 9 '18 at 19:58
fatimaaafatimaaa
12
$endgroup$
add a comment |
$begingroup$
I would like to find this probability: $Pr(Z<2^{2R})$ for $R>0$.
So, I try several ways and finally decide to find pdf of Z.
If $X$ and $Y$ are independent and exponentially distributed with parameter $lambda_1$ and $lambda_2$, respectively, and $a_i>0$ for $i=1,...,6$, which is the pdf of $Z$? Where $Z$ is given by
$$Z=frac{(a_1X+a_2)(a_3Y+a_5)}{(a_3X+a_4)(a_1Y+a_6)} $$
any idea?
First,considering $Z=W_{1}W_{2}$, I calculate pdf of $W1=frac{(a_1X+a_2)}{(a_3X+a_4)}=frac{W_{11}}{W_{12}}$ as:
$f_{W_{11}}(w_{11})=frac{1}{a_{1}}f_{X}(frac{w_{11}-a_{2}}{a_{1}})$ for $w_{11}>a_{2}$
$f_{W_{12}}(w_{12})=frac{1}{a_{3}}f_{X}(frac{w_{12}-a_{4}}{a_{3}})$ for $w_{12}>a_{4}$
$F_{W_{1}}(W_{1})=Pr(W1=frac{W_{11}}{W_{12}}<w_{1})=int_{a_{4}}^{infty}int_{a_{2}}^{ww_{2}} f_{w_{11}}(w_{11})f_{w_{12}}(w_{12}), dw_{11}dw_{12}=1-frac{{{e}^{{{lambda }_{1}}(frac{{{a}_{2}}}{{{a}_{1}}}+frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{lambda }_{1}}(frac{{{a}_{4}}}{{{a}_{3}}}+frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{(1+frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})}$
If we derive from the above, pdf of $W_{1}$ is achieved as:
$frac{a_{3}{{e}^{{{lambda }_{1}}(frac{{{a}_{2}}}{{{a}_{1}}}+frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{lambda }_{1}}(frac{{{a}_{4}}}{{{a}_{3}}}+frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{a_{1}(1+frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})^{2}}+frac{lambda_{1} a_{4}{{e}^{{{lambda }_{1}}(frac{{{a}_{2}}}{{{a}_{1}}}+frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{lambda }_{1}}(frac{{{a}_{4}}}{{{a}_{3}}}+frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{a_{1}(1+frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})}$
But, I have a problem: I know that integral of $f_{W_{11}}(W_{11})$ over whole interval equal to $1$. What are the integral bounds?(I need these bounds for calculating pdf of product of $W_{1}W_{2}$)
probability probability-distributions random-variables
asked Dec 9 '18 at 19:58
fatimaaafatimaaa
12
$endgroup$
I would like to find this probability: $Pr(Z<2^{2R})$ for $R>0$.
So, I try several ways and finally decide to find pdf of Z.
If $X$ and $Y$ are independent and exponentially distributed with parameter $lambda_1$ and $lambda_2$, respectively, and $a_i>0$ for $i=1,...,6$, which is the pdf of $Z$? Where $Z$ is given by
$$Z=frac{(a_1X+a_2)(a_3Y+a_5)}{(a_3X+a_4)(a_1Y+a_6)} $$
any idea?
First,considering $Z=W_{1}W_{2}$, I calculate pdf of $W1=frac{(a_1X+a_2)}{(a_3X+a_4)}=frac{W_{11}}{W_{12}}$ as:
$f_{W_{11}}(w_{11})=frac{1}{a_{1}}f_{X}(frac{w_{11}-a_{2}}{a_{1}})$ for $w_{11}>a_{2}$
$f_{W_{12}}(w_{12})=frac{1}{a_{3}}f_{X}(frac{w_{12}-a_{4}}{a_{3}})$ for $w_{12}>a_{4}$
$F_{W_{1}}(W_{1})=Pr(W1=frac{W_{11}}{W_{12}}<w_{1})=int_{a_{4}}^{infty}int_{a_{2}}^{ww_{2}} f_{w_{11}}(w_{11})f_{w_{12}}(w_{12}), dw_{11}dw_{12}=1-frac{{{e}^{{{lambda }_{1}}(frac{{{a}_{2}}}{{{a}_{1}}}+frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{lambda }_{1}}(frac{{{a}_{4}}}{{{a}_{3}}}+frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{(1+frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})}$
If we derive from the above, pdf of $W_{1}$ is achieved as:
$frac{a_{3}{{e}^{{{lambda }_{1}}(frac{{{a}_{2}}}{{{a}_{1}}}+frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{lambda }_{1}}(frac{{{a}_{4}}}{{{a}_{3}}}+frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{a_{1}(1+frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})^{2}}+frac{lambda_{1} a_{4}{{e}^{{{lambda }_{1}}(frac{{{a}_{2}}}{{{a}_{1}}}+frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{lambda }_{1}}(frac{{{a}_{4}}}{{{a}_{3}}}+frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{a_{1}(1+frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})}$
But, I have a problem: I know that integral of $f_{W_{11}}(W_{11})$ over whole interval equal to $1$. What are the integral bounds?(I need these bounds for calculating pdf of product of $W_{1}W_{2}$)
probability probability-distributions random-variables
probability probability-distributions random-variables
asked Dec 9 '18 at 19:58
fatimaaafatimaaa
12
asked Dec 9 '18 at 19:58
fatimaaafatimaaa
12
edited Dec 10 '18 at 16:17
fatimaaa
asked Dec 9 '18 at 19:58
fatimaaafatimaaa
12
asked Dec 9 '18 at 19:58
fatimaaafatimaaa
12
asked Dec 9 '18 at 19:58
fatimaaafatimaaa
12
12
add a comment |
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