Computing the joint CDF of $(X,X^3)$ with $X$ exponential












1












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1


1



$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 9:04









Did

248k23224462




248k23224462










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


















  • $begingroup$
    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










1 Answer
1






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3












$begingroup$

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
$$
F(x,y)=1-e^{-lambdamin{x,y^{1/3}}}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you. Very easy to understand.
    $endgroup$
    – Donald Mayer
    Dec 17 '18 at 16:25











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

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
$$
F(x,y)=1-e^{-lambdamin{x,y^{1/3}}}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you. Very easy to understand.
    $endgroup$
    – Donald Mayer
    Dec 17 '18 at 16:25
















3












$begingroup$

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
$$
F(x,y)=1-e^{-lambdamin{x,y^{1/3}}}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you. Very easy to understand.
    $endgroup$
    – Donald Mayer
    Dec 17 '18 at 16:25














3












3








3





$begingroup$

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
$$
F(x,y)=1-e^{-lambdamin{x,y^{1/3}}}
$$






share|cite|improve this answer











$endgroup$



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
$$
F(x,y)=1-e^{-lambdamin{x,y^{1/3}}}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 16:40

























answered Dec 17 '18 at 8:57









DidDid

248k23224462




248k23224462












  • $begingroup$
    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




$begingroup$
Thank you. Very easy to understand.
$endgroup$
– Donald Mayer
Dec 17 '18 at 16:25


















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