Mixing
CDF of $Z=X_1+max{X_2,,X_3}$
$begingroup$
Let $X_1, X_2$ and $X_3$ be expontial random varibles with paramtre $beta_1, beta_2$ and $beta_3$ respectivly.
The PDF and CDF of $X_i$ for $xgeq 0$ are $f_{X_i}(x)$ and $F_{X_i}(x)$ given by
begin{align}
f_{X_i}(x)&=b_ie^{-x beta_i} \
F_{X_i}(x)&=1-e^{-x beta_i}.
end{align}
I would like to find the CDF of random varible $Z$ define by
$$Z=X_1+max{X2,X3}$$ for difrent parametre.
What is the CDF if $beta_i=beta$ for all $iin{1,2,3}$?.
Thanks.
random-variables exponential-distribution
edited Feb 2 at 20:02
max_zorn
3,44061429
asked Feb 2 at 13:34
Mokh Tar BouMokh Tar Bou
607
$endgroup$
add a comment |
$begingroup$
Let $X_1, X_2$ and $X_3$ be expontial random varibles with paramtre $beta_1, beta_2$ and $beta_3$ respectivly.
The PDF and CDF of $X_i$ for $xgeq 0$ are $f_{X_i}(x)$ and $F_{X_i}(x)$ given by
begin{align}
f_{X_i}(x)&=b_ie^{-x beta_i} \
F_{X_i}(x)&=1-e^{-x beta_i}.
end{align}
I would like to find the CDF of random varible $Z$ define by
$$Z=X_1+max{X2,X3}$$ for difrent parametre.
What is the CDF if $beta_i=beta$ for all $iin{1,2,3}$?.
Thanks.
random-variables exponential-distribution
edited Feb 2 at 20:02
max_zorn
3,44061429
asked Feb 2 at 13:34
Mokh Tar BouMokh Tar Bou
607
$endgroup$
$begingroup$
What is stopping you here?
$endgroup$
– Did
Feb 2 at 17:00
add a comment |
$begingroup$
Let $X_1, X_2$ and $X_3$ be expontial random varibles with paramtre $beta_1, beta_2$ and $beta_3$ respectivly.
The PDF and CDF of $X_i$ for $xgeq 0$ are $f_{X_i}(x)$ and $F_{X_i}(x)$ given by
begin{align}
f_{X_i}(x)&=b_ie^{-x beta_i} \
F_{X_i}(x)&=1-e^{-x beta_i}.
end{align}
I would like to find the CDF of random varible $Z$ define by
$$Z=X_1+max{X2,X3}$$ for difrent parametre.
What is the CDF if $beta_i=beta$ for all $iin{1,2,3}$?.
Thanks.
random-variables exponential-distribution
edited Feb 2 at 20:02
max_zorn
3,44061429
asked Feb 2 at 13:34
Mokh Tar BouMokh Tar Bou
607
$endgroup$
Let $X_1, X_2$ and $X_3$ be expontial random varibles with paramtre $beta_1, beta_2$ and $beta_3$ respectivly.
The PDF and CDF of $X_i$ for $xgeq 0$ are $f_{X_i}(x)$ and $F_{X_i}(x)$ given by
begin{align}
f_{X_i}(x)&=b_ie^{-x beta_i} \
F_{X_i}(x)&=1-e^{-x beta_i}.
end{align}
I would like to find the CDF of random varible $Z$ define by
$$Z=X_1+max{X2,X3}$$ for difrent parametre.
What is the CDF if $beta_i=beta$ for all $iin{1,2,3}$?.
Thanks.
random-variables exponential-distribution
random-variables exponential-distribution
edited Feb 2 at 20:02
max_zorn
3,44061429
asked Feb 2 at 13:34
Mokh Tar BouMokh Tar Bou
607
edited Feb 2 at 20:02
max_zorn
3,44061429
asked Feb 2 at 13:34
Mokh Tar BouMokh Tar Bou
607
edited Feb 2 at 20:02
max_zorn
3,44061429
edited Feb 2 at 20:02
max_zorn
3,44061429
edited Feb 2 at 20:02
max_zorn
3,44061429
3,44061429
asked Feb 2 at 13:34
Mokh Tar BouMokh Tar Bou
607
asked Feb 2 at 13:34
Mokh Tar BouMokh Tar Bou
607
asked Feb 2 at 13:34
Mokh Tar BouMokh Tar Bou
607
607
$begingroup$
What is stopping you here?
$endgroup$
– Did
Feb 2 at 17:00
add a comment |
$begingroup$
What is stopping you here?
$endgroup$
– Did
Feb 2 at 17:00
$begingroup$
What is stopping you here?
$endgroup$
– Did
Feb 2 at 17:00
$begingroup$
What is stopping you here?
$endgroup$
– Did
Feb 2 at 17:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is the type of problem that is not well-suited to manual solution, and which can benefit considerably from the assistance of a computer algebra system.
If $X_1$, $X_2$ and $X_3$ are independent $text{Exponential}(b_i)$ random variables, then the joint pdf $f(x_1,x_2,x_3)$ is:
We seek the cdf of $Z = X_1 + max(X_2,X_3)$, namely $P(Z<z)$.
This can be obtained immediately as:
which returns the cdf as:
... where I have used the Prob function from the mathStatica add-on to Mathematica to help automate the calculation (and as disclosure, of which I am one of the authors).
Identical parameters
In the case where the $b_i$ are identical, the set-up is identical: simply replace each $b_i$ with $b$, which yields a much more elegant solution for the cdf:
$$F(z) = 2 e^{-b z} (sinh (b z)-b z) quad quad text{ for } z > 0$$
Monte Carlo check
It is always a good idea to test symbolic work by alternative methods. Here is a quick comparison of the theoretical pdf (red dashed curve) with the empirical pdf (squiggly blue curve - generated by Monte Carlo) when $b_1 = 10$, $b_2 = 0.4$, and $b_3 = 7$:
All looks good.
answered Feb 2 at 16:21
wolfieswolfies
4,2692923
$endgroup$
add a comment |
$begingroup$
Let $Y:=max{X_2,,X_3}$. Since for $yge 0$ $$P(Yle y)=P(X_2le yland X_3le y)=1-exp -ybeta_2-exp -ybeta_3+exp -y(beta_2+beta_3),$$ the pdf of $Y$ is $$f_Y(y):=beta_2exp -ybeta_2+beta_3exp -ybeta_3-(beta_2+beta_3)exp -y(beta_2+beta_3).$$Since $X_1$ has pdf $f_X(x):=beta_1exp -xbeta_1$ for $xge 0$, $Z$ has pdf$$int_0^z f_X(z-y)f_Y(y)dy,$$and this you can easily compute and integrate.
answered Feb 2 at 13:53
J.G.J.G.
33.5k23252
$endgroup$
$begingroup$
Ok but I am loking for CDF not PDF?, thank
$endgroup$
– Mokh Tar Bou
Feb 2 at 13:58
$begingroup$
@MokhTarBou I know; I told you to (i) compute the integral, which is a PDF, then (ii) integrate that function to get the CDF.
$endgroup$
– J.G.
Feb 2 at 14:09
$begingroup$
ok I wil try but how I cheek if it the correct anser?
$endgroup$
– Mokh Tar Bou
Feb 2 at 14:12
$begingroup$
I am not sure I would agree that the integral is easy to compute, or that the derivation of the pdf of the max is correct.
$endgroup$
– wolfies
Feb 2 at 15:07
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is the type of problem that is not well-suited to manual solution, and which can benefit considerably from the assistance of a computer algebra system.
If $X_1$, $X_2$ and $X_3$ are independent $text{Exponential}(b_i)$ random variables, then the joint pdf $f(x_1,x_2,x_3)$ is:
We seek the cdf of $Z = X_1 + max(X_2,X_3)$, namely $P(Z<z)$.
This can be obtained immediately as:
which returns the cdf as:
... where I have used the Prob function from the mathStatica add-on to Mathematica to help automate the calculation (and as disclosure, of which I am one of the authors).
Identical parameters
In the case where the $b_i$ are identical, the set-up is identical: simply replace each $b_i$ with $b$, which yields a much more elegant solution for the cdf:
$$F(z) = 2 e^{-b z} (sinh (b z)-b z) quad quad text{ for } z > 0$$
Monte Carlo check
It is always a good idea to test symbolic work by alternative methods. Here is a quick comparison of the theoretical pdf (red dashed curve) with the empirical pdf (squiggly blue curve - generated by Monte Carlo) when $b_1 = 10$, $b_2 = 0.4$, and $b_3 = 7$:
All looks good.
answered Feb 2 at 16:21
wolfieswolfies
4,2692923
$endgroup$
add a comment |
$begingroup$
This is the type of problem that is not well-suited to manual solution, and which can benefit considerably from the assistance of a computer algebra system.
If $X_1$, $X_2$ and $X_3$ are independent $text{Exponential}(b_i)$ random variables, then the joint pdf $f(x_1,x_2,x_3)$ is:
We seek the cdf of $Z = X_1 + max(X_2,X_3)$, namely $P(Z<z)$.
This can be obtained immediately as:
which returns the cdf as:
... where I have used the Prob function from the mathStatica add-on to Mathematica to help automate the calculation (and as disclosure, of which I am one of the authors).
Identical parameters
In the case where the $b_i$ are identical, the set-up is identical: simply replace each $b_i$ with $b$, which yields a much more elegant solution for the cdf:
$$F(z) = 2 e^{-b z} (sinh (b z)-b z) quad quad text{ for } z > 0$$
Monte Carlo check
It is always a good idea to test symbolic work by alternative methods. Here is a quick comparison of the theoretical pdf (red dashed curve) with the empirical pdf (squiggly blue curve - generated by Monte Carlo) when $b_1 = 10$, $b_2 = 0.4$, and $b_3 = 7$:
All looks good.
answered Feb 2 at 16:21
wolfieswolfies
4,2692923
$endgroup$
add a comment |
$begingroup$
This is the type of problem that is not well-suited to manual solution, and which can benefit considerably from the assistance of a computer algebra system.
If $X_1$, $X_2$ and $X_3$ are independent $text{Exponential}(b_i)$ random variables, then the joint pdf $f(x_1,x_2,x_3)$ is:
We seek the cdf of $Z = X_1 + max(X_2,X_3)$, namely $P(Z<z)$.
This can be obtained immediately as:
which returns the cdf as:
... where I have used the Prob function from the mathStatica add-on to Mathematica to help automate the calculation (and as disclosure, of which I am one of the authors).
Identical parameters
In the case where the $b_i$ are identical, the set-up is identical: simply replace each $b_i$ with $b$, which yields a much more elegant solution for the cdf:
$$F(z) = 2 e^{-b z} (sinh (b z)-b z) quad quad text{ for } z > 0$$
Monte Carlo check
It is always a good idea to test symbolic work by alternative methods. Here is a quick comparison of the theoretical pdf (red dashed curve) with the empirical pdf (squiggly blue curve - generated by Monte Carlo) when $b_1 = 10$, $b_2 = 0.4$, and $b_3 = 7$:
All looks good.
answered Feb 2 at 16:21
wolfieswolfies
4,2692923
$endgroup$
This is the type of problem that is not well-suited to manual solution, and which can benefit considerably from the assistance of a computer algebra system.
If $X_1$, $X_2$ and $X_3$ are independent $text{Exponential}(b_i)$ random variables, then the joint pdf $f(x_1,x_2,x_3)$ is:
We seek the cdf of $Z = X_1 + max(X_2,X_3)$, namely $P(Z<z)$.
This can be obtained immediately as:
which returns the cdf as:
... where I have used the Prob function from the mathStatica add-on to Mathematica to help automate the calculation (and as disclosure, of which I am one of the authors).
Identical parameters
In the case where the $b_i$ are identical, the set-up is identical: simply replace each $b_i$ with $b$, which yields a much more elegant solution for the cdf:
$$F(z) = 2 e^{-b z} (sinh (b z)-b z) quad quad text{ for } z > 0$$
Monte Carlo check
It is always a good idea to test symbolic work by alternative methods. Here is a quick comparison of the theoretical pdf (red dashed curve) with the empirical pdf (squiggly blue curve - generated by Monte Carlo) when $b_1 = 10$, $b_2 = 0.4$, and $b_3 = 7$:
All looks good.
answered Feb 2 at 16:21
wolfieswolfies
4,2692923
answered Feb 2 at 16:21
wolfieswolfies
4,2692923
answered Feb 2 at 16:21
wolfieswolfies
4,2692923
answered Feb 2 at 16:21
wolfieswolfies
4,2692923
4,2692923
add a comment |
add a comment |
$begingroup$
Let $Y:=max{X_2,,X_3}$. Since for $yge 0$ $$P(Yle y)=P(X_2le yland X_3le y)=1-exp -ybeta_2-exp -ybeta_3+exp -y(beta_2+beta_3),$$ the pdf of $Y$ is $$f_Y(y):=beta_2exp -ybeta_2+beta_3exp -ybeta_3-(beta_2+beta_3)exp -y(beta_2+beta_3).$$Since $X_1$ has pdf $f_X(x):=beta_1exp -xbeta_1$ for $xge 0$, $Z$ has pdf$$int_0^z f_X(z-y)f_Y(y)dy,$$and this you can easily compute and integrate.
answered Feb 2 at 13:53
J.G.J.G.
33.5k23252
$endgroup$
$begingroup$
Ok but I am loking for CDF not PDF?, thank
$endgroup$
– Mokh Tar Bou
Feb 2 at 13:58
$begingroup$
@MokhTarBou I know; I told you to (i) compute the integral, which is a PDF, then (ii) integrate that function to get the CDF.
$endgroup$
– J.G.
Feb 2 at 14:09
$begingroup$
ok I wil try but how I cheek if it the correct anser?
$endgroup$
– Mokh Tar Bou
Feb 2 at 14:12
$begingroup$
I am not sure I would agree that the integral is easy to compute, or that the derivation of the pdf of the max is correct.
$endgroup$
– wolfies
Feb 2 at 15:07
add a comment |
$begingroup$
Let $Y:=max{X_2,,X_3}$. Since for $yge 0$ $$P(Yle y)=P(X_2le yland X_3le y)=1-exp -ybeta_2-exp -ybeta_3+exp -y(beta_2+beta_3),$$ the pdf of $Y$ is $$f_Y(y):=beta_2exp -ybeta_2+beta_3exp -ybeta_3-(beta_2+beta_3)exp -y(beta_2+beta_3).$$Since $X_1$ has pdf $f_X(x):=beta_1exp -xbeta_1$ for $xge 0$, $Z$ has pdf$$int_0^z f_X(z-y)f_Y(y)dy,$$and this you can easily compute and integrate.
answered Feb 2 at 13:53
J.G.J.G.
33.5k23252
$endgroup$
$begingroup$
Ok but I am loking for CDF not PDF?, thank
$endgroup$
– Mokh Tar Bou
Feb 2 at 13:58
$begingroup$
@MokhTarBou I know; I told you to (i) compute the integral, which is a PDF, then (ii) integrate that function to get the CDF.
$endgroup$
– J.G.
Feb 2 at 14:09
$begingroup$
ok I wil try but how I cheek if it the correct anser?
$endgroup$
– Mokh Tar Bou
Feb 2 at 14:12
$begingroup$
I am not sure I would agree that the integral is easy to compute, or that the derivation of the pdf of the max is correct.
$endgroup$
– wolfies
Feb 2 at 15:07
add a comment |
$begingroup$
Let $Y:=max{X_2,,X_3}$. Since for $yge 0$ $$P(Yle y)=P(X_2le yland X_3le y)=1-exp -ybeta_2-exp -ybeta_3+exp -y(beta_2+beta_3),$$ the pdf of $Y$ is $$f_Y(y):=beta_2exp -ybeta_2+beta_3exp -ybeta_3-(beta_2+beta_3)exp -y(beta_2+beta_3).$$Since $X_1$ has pdf $f_X(x):=beta_1exp -xbeta_1$ for $xge 0$, $Z$ has pdf$$int_0^z f_X(z-y)f_Y(y)dy,$$and this you can easily compute and integrate.
answered Feb 2 at 13:53
J.G.J.G.
33.5k23252
$endgroup$
Let $Y:=max{X_2,,X_3}$. Since for $yge 0$ $$P(Yle y)=P(X_2le yland X_3le y)=1-exp -ybeta_2-exp -ybeta_3+exp -y(beta_2+beta_3),$$ the pdf of $Y$ is $$f_Y(y):=beta_2exp -ybeta_2+beta_3exp -ybeta_3-(beta_2+beta_3)exp -y(beta_2+beta_3).$$Since $X_1$ has pdf $f_X(x):=beta_1exp -xbeta_1$ for $xge 0$, $Z$ has pdf$$int_0^z f_X(z-y)f_Y(y)dy,$$and this you can easily compute and integrate.
answered Feb 2 at 13:53
J.G.J.G.
33.5k23252
answered Feb 2 at 13:53
J.G.J.G.
33.5k23252
answered Feb 2 at 13:53
J.G.J.G.
33.5k23252
answered Feb 2 at 13:53
J.G.J.G.
33.5k23252
33.5k23252
$begingroup$
Ok but I am loking for CDF not PDF?, thank
$endgroup$
– Mokh Tar Bou
Feb 2 at 13:58
$begingroup$
@MokhTarBou I know; I told you to (i) compute the integral, which is a PDF, then (ii) integrate that function to get the CDF.
$endgroup$
– J.G.
Feb 2 at 14:09
$begingroup$
ok I wil try but how I cheek if it the correct anser?
$endgroup$
– Mokh Tar Bou
Feb 2 at 14:12
$begingroup$
I am not sure I would agree that the integral is easy to compute, or that the derivation of the pdf of the max is correct.
$endgroup$
– wolfies
Feb 2 at 15:07
add a comment |
$begingroup$
Ok but I am loking for CDF not PDF?, thank
$endgroup$
– Mokh Tar Bou
Feb 2 at 13:58
$begingroup$
@MokhTarBou I know; I told you to (i) compute the integral, which is a PDF, then (ii) integrate that function to get the CDF.
$endgroup$
– J.G.
Feb 2 at 14:09
$begingroup$
ok I wil try but how I cheek if it the correct anser?
$endgroup$
– Mokh Tar Bou
Feb 2 at 14:12
$begingroup$
I am not sure I would agree that the integral is easy to compute, or that the derivation of the pdf of the max is correct.
$endgroup$
– wolfies
Feb 2 at 15:07
$begingroup$
Ok but I am loking for CDF not PDF?, thank
$endgroup$
– Mokh Tar Bou
Feb 2 at 13:58
$begingroup$
Ok but I am loking for CDF not PDF?, thank
$endgroup$
– Mokh Tar Bou
Feb 2 at 13:58
$begingroup$
@MokhTarBou I know; I told you to (i) compute the integral, which is a PDF, then (ii) integrate that function to get the CDF.
$endgroup$
– J.G.
Feb 2 at 14:09
$begingroup$
@MokhTarBou I know; I told you to (i) compute the integral, which is a PDF, then (ii) integrate that function to get the CDF.
$endgroup$
– J.G.
Feb 2 at 14:09
$begingroup$
ok I wil try but how I cheek if it the correct anser?
$endgroup$
– Mokh Tar Bou
Feb 2 at 14:12
$begingroup$
ok I wil try but how I cheek if it the correct anser?
$endgroup$
– Mokh Tar Bou
Feb 2 at 14:12
$begingroup$
I am not sure I would agree that the integral is easy to compute, or that the derivation of the pdf of the max is correct.
$endgroup$
– wolfies
Feb 2 at 15:07
$begingroup$
I am not sure I would agree that the integral is easy to compute, or that the derivation of the pdf of the max is correct.
$endgroup$
– wolfies
Feb 2 at 15:07
add a comment |
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What is stopping you here?
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– Did
Feb 2 at 17:00