Computing the joint CDF of $(X,X^3)$ with $X$ exponential
$begingroup$
Let $Xsim text{Exponential}(lambda)$, and let $Y=X^3$. Compute the joint CDF $F_{X,Y}$ of $(X,Y)$.
With my current understanding, I can only come up with using the one dimensional change in variable where I set $F_Y(y) = P(Y leq y) = P((lambda e^{-lambda x})^3leq y)$, then isolating for y, and then apply the transformation.
I am not sure if this is even a valid solution.
probability-theory probability-distributions
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|
show 7 more comments
$begingroup$
Let $Xsim text{Exponential}(lambda)$, and let $Y=X^3$. Compute the joint CDF $F_{X,Y}$ of $(X,Y)$.
With my current understanding, I can only come up with using the one dimensional change in variable where I set $F_Y(y) = P(Y leq y) = P((lambda e^{-lambda x})^3leq y)$, then isolating for y, and then apply the transformation.
I am not sure if this is even a valid solution.
probability-theory probability-distributions
$endgroup$
$begingroup$
For transformation $X=Y^{frac{1}{3}}$ Apply $g(y)=f(x)|frac{dx}{dy}|$ Note that this formula works only when your g(x) is monotone .
$endgroup$
– Daman deep
Dec 17 '18 at 6:03
$begingroup$
$Y$ and $X$ are totally dependant. The pdf should include something like $delta(y-x^ 3)$, the cdf something equivalent
$endgroup$
– Damien
Dec 17 '18 at 6:28
$begingroup$
You may use $min(y, x^ 3)$
$endgroup$
– Damien
Dec 17 '18 at 6:45
$begingroup$
@Damandeep That's not something that was taught in my course. Thanks for the suggestion though.
$endgroup$
– Donald Mayer
Dec 17 '18 at 6:50
1
$begingroup$
@Did $F_Y(y) = P(Y leq y) = P((lambda e^{-lambda x})^3leq y)$ His substitution of $Y$ is wrong .
$endgroup$
– Daman deep
Dec 17 '18 at 9:02
|
show 7 more comments
$begingroup$
Let $Xsim text{Exponential}(lambda)$, and let $Y=X^3$. Compute the joint CDF $F_{X,Y}$ of $(X,Y)$.
With my current understanding, I can only come up with using the one dimensional change in variable where I set $F_Y(y) = P(Y leq y) = P((lambda e^{-lambda x})^3leq y)$, then isolating for y, and then apply the transformation.
I am not sure if this is even a valid solution.
probability-theory probability-distributions
$endgroup$
Let $Xsim text{Exponential}(lambda)$, and let $Y=X^3$. Compute the joint CDF $F_{X,Y}$ of $(X,Y)$.
With my current understanding, I can only come up with using the one dimensional change in variable where I set $F_Y(y) = P(Y leq y) = P((lambda e^{-lambda x})^3leq y)$, then isolating for y, and then apply the transformation.
I am not sure if this is even a valid solution.
probability-theory probability-distributions
probability-theory probability-distributions
edited Dec 17 '18 at 9:04
Did
248k23224463
248k23224463
asked Dec 17 '18 at 5:56
Donald MayerDonald Mayer
266
266
$begingroup$
For transformation $X=Y^{frac{1}{3}}$ Apply $g(y)=f(x)|frac{dx}{dy}|$ Note that this formula works only when your g(x) is monotone .
$endgroup$
– Daman deep
Dec 17 '18 at 6:03
$begingroup$
$Y$ and $X$ are totally dependant. The pdf should include something like $delta(y-x^ 3)$, the cdf something equivalent
$endgroup$
– Damien
Dec 17 '18 at 6:28
$begingroup$
You may use $min(y, x^ 3)$
$endgroup$
– Damien
Dec 17 '18 at 6:45
$begingroup$
@Damandeep That's not something that was taught in my course. Thanks for the suggestion though.
$endgroup$
– Donald Mayer
Dec 17 '18 at 6:50
1
$begingroup$
@Did $F_Y(y) = P(Y leq y) = P((lambda e^{-lambda x})^3leq y)$ His substitution of $Y$ is wrong .
$endgroup$
– Daman deep
Dec 17 '18 at 9:02
|
show 7 more comments
$begingroup$
For transformation $X=Y^{frac{1}{3}}$ Apply $g(y)=f(x)|frac{dx}{dy}|$ Note that this formula works only when your g(x) is monotone .
$endgroup$
– Daman deep
Dec 17 '18 at 6:03
$begingroup$
$Y$ and $X$ are totally dependant. The pdf should include something like $delta(y-x^ 3)$, the cdf something equivalent
$endgroup$
– Damien
Dec 17 '18 at 6:28
$begingroup$
You may use $min(y, x^ 3)$
$endgroup$
– Damien
Dec 17 '18 at 6:45
$begingroup$
@Damandeep That's not something that was taught in my course. Thanks for the suggestion though.
$endgroup$
– Donald Mayer
Dec 17 '18 at 6:50
1
$begingroup$
@Did $F_Y(y) = P(Y leq y) = P((lambda e^{-lambda x})^3leq y)$ His substitution of $Y$ is wrong .
$endgroup$
– Daman deep
Dec 17 '18 at 9:02
$begingroup$
For transformation $X=Y^{frac{1}{3}}$ Apply $g(y)=f(x)|frac{dx}{dy}|$ Note that this formula works only when your g(x) is monotone .
$endgroup$
– Daman deep
Dec 17 '18 at 6:03
$begingroup$
For transformation $X=Y^{frac{1}{3}}$ Apply $g(y)=f(x)|frac{dx}{dy}|$ Note that this formula works only when your g(x) is monotone .
$endgroup$
– Daman deep
Dec 17 '18 at 6:03
$begingroup$
$Y$ and $X$ are totally dependant. The pdf should include something like $delta(y-x^ 3)$, the cdf something equivalent
$endgroup$
– Damien
Dec 17 '18 at 6:28
$begingroup$
$Y$ and $X$ are totally dependant. The pdf should include something like $delta(y-x^ 3)$, the cdf something equivalent
$endgroup$
– Damien
Dec 17 '18 at 6:28
$begingroup$
You may use $min(y, x^ 3)$
$endgroup$
– Damien
Dec 17 '18 at 6:45
$begingroup$
You may use $min(y, x^ 3)$
$endgroup$
– Damien
Dec 17 '18 at 6:45
$begingroup$
@Damandeep That's not something that was taught in my course. Thanks for the suggestion though.
$endgroup$
– Donald Mayer
Dec 17 '18 at 6:50
$begingroup$
@Damandeep That's not something that was taught in my course. Thanks for the suggestion though.
$endgroup$
– Donald Mayer
Dec 17 '18 at 6:50
1
1
$begingroup$
@Did $F_Y(y) = P(Y leq y) = P((lambda e^{-lambda x})^3leq y)$ His substitution of $Y$ is wrong .
$endgroup$
– Daman deep
Dec 17 '18 at 9:02
$begingroup$
@Did $F_Y(y) = P(Y leq y) = P((lambda e^{-lambda x})^3leq y)$ His substitution of $Y$ is wrong .
$endgroup$
– Daman deep
Dec 17 '18 at 9:02
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The PDF of $(X,Y)$ does not exist since the support of the distribution of $(X,Y)$ is a subset of the set ${(x,y)inmathbb R^2,;,y=x^3}$, whose measure is zero.
To compute the CDF $F$ of $(X,Y)$, note that $${Xleqslant x,Yleqslant y}={Xleqslant x,X^3leqslant y}={Xleqslant x,Xleqslant y^{1/3}}={Xleqslantmin{x,y^{1/3}}}$$ In terms of the CDF $F_X$ of $X$, one gets
$$F(x,y)=F_X(min{x,y^{1/3}})$$
In the present case, if $x$ or $y$ is negative, then $F(x,y)=0$, and, if $x$ and $y$ are nonnegative, then
$$
F(x,y)=1-e^{-lambdamin{x,y^{1/3}}}
$$
$endgroup$
$begingroup$
Thank you. Very easy to understand.
$endgroup$
– Donald Mayer
Dec 17 '18 at 16:25
add a comment |
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1 Answer
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$begingroup$
The PDF of $(X,Y)$ does not exist since the support of the distribution of $(X,Y)$ is a subset of the set ${(x,y)inmathbb R^2,;,y=x^3}$, whose measure is zero.
To compute the CDF $F$ of $(X,Y)$, note that $${Xleqslant x,Yleqslant y}={Xleqslant x,X^3leqslant y}={Xleqslant x,Xleqslant y^{1/3}}={Xleqslantmin{x,y^{1/3}}}$$ In terms of the CDF $F_X$ of $X$, one gets
$$F(x,y)=F_X(min{x,y^{1/3}})$$
In the present case, if $x$ or $y$ is negative, then $F(x,y)=0$, and, if $x$ and $y$ are nonnegative, then
$$
F(x,y)=1-e^{-lambdamin{x,y^{1/3}}}
$$
$endgroup$
$begingroup$
Thank you. Very easy to understand.
$endgroup$
– Donald Mayer
Dec 17 '18 at 16:25
add a comment |
$begingroup$
The PDF of $(X,Y)$ does not exist since the support of the distribution of $(X,Y)$ is a subset of the set ${(x,y)inmathbb R^2,;,y=x^3}$, whose measure is zero.
To compute the CDF $F$ of $(X,Y)$, note that $${Xleqslant x,Yleqslant y}={Xleqslant x,X^3leqslant y}={Xleqslant x,Xleqslant y^{1/3}}={Xleqslantmin{x,y^{1/3}}}$$ In terms of the CDF $F_X$ of $X$, one gets
$$F(x,y)=F_X(min{x,y^{1/3}})$$
In the present case, if $x$ or $y$ is negative, then $F(x,y)=0$, and, if $x$ and $y$ are nonnegative, then
$$
F(x,y)=1-e^{-lambdamin{x,y^{1/3}}}
$$
$endgroup$
$begingroup$
Thank you. Very easy to understand.
$endgroup$
– Donald Mayer
Dec 17 '18 at 16:25
add a comment |
$begingroup$
The PDF of $(X,Y)$ does not exist since the support of the distribution of $(X,Y)$ is a subset of the set ${(x,y)inmathbb R^2,;,y=x^3}$, whose measure is zero.
To compute the CDF $F$ of $(X,Y)$, note that $${Xleqslant x,Yleqslant y}={Xleqslant x,X^3leqslant y}={Xleqslant x,Xleqslant y^{1/3}}={Xleqslantmin{x,y^{1/3}}}$$ In terms of the CDF $F_X$ of $X$, one gets
$$F(x,y)=F_X(min{x,y^{1/3}})$$
In the present case, if $x$ or $y$ is negative, then $F(x,y)=0$, and, if $x$ and $y$ are nonnegative, then
$$
F(x,y)=1-e^{-lambdamin{x,y^{1/3}}}
$$
$endgroup$
The PDF of $(X,Y)$ does not exist since the support of the distribution of $(X,Y)$ is a subset of the set ${(x,y)inmathbb R^2,;,y=x^3}$, whose measure is zero.
To compute the CDF $F$ of $(X,Y)$, note that $${Xleqslant x,Yleqslant y}={Xleqslant x,X^3leqslant y}={Xleqslant x,Xleqslant y^{1/3}}={Xleqslantmin{x,y^{1/3}}}$$ In terms of the CDF $F_X$ of $X$, one gets
$$F(x,y)=F_X(min{x,y^{1/3}})$$
In the present case, if $x$ or $y$ is negative, then $F(x,y)=0$, and, if $x$ and $y$ are nonnegative, then
$$
F(x,y)=1-e^{-lambdamin{x,y^{1/3}}}
$$
edited Dec 17 '18 at 16:40
answered Dec 17 '18 at 8:57
DidDid
248k23224463
248k23224463
$begingroup$
Thank you. Very easy to understand.
$endgroup$
– Donald Mayer
Dec 17 '18 at 16:25
add a comment |
$begingroup$
Thank you. Very easy to understand.
$endgroup$
– Donald Mayer
Dec 17 '18 at 16:25
$begingroup$
Thank you. Very easy to understand.
$endgroup$
– Donald Mayer
Dec 17 '18 at 16:25
$begingroup$
Thank you. Very easy to understand.
$endgroup$
– Donald Mayer
Dec 17 '18 at 16:25
add a comment |
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$begingroup$
For transformation $X=Y^{frac{1}{3}}$ Apply $g(y)=f(x)|frac{dx}{dy}|$ Note that this formula works only when your g(x) is monotone .
$endgroup$
– Daman deep
Dec 17 '18 at 6:03
$begingroup$
$Y$ and $X$ are totally dependant. The pdf should include something like $delta(y-x^ 3)$, the cdf something equivalent
$endgroup$
– Damien
Dec 17 '18 at 6:28
$begingroup$
You may use $min(y, x^ 3)$
$endgroup$
– Damien
Dec 17 '18 at 6:45
$begingroup$
@Damandeep That's not something that was taught in my course. Thanks for the suggestion though.
$endgroup$
– Donald Mayer
Dec 17 '18 at 6:50
1
$begingroup$
@Did $F_Y(y) = P(Y leq y) = P((lambda e^{-lambda x})^3leq y)$ His substitution of $Y$ is wrong .
$endgroup$
– Daman deep
Dec 17 '18 at 9:02