How to find bounds of integration in finding CDF of $XY$
Suppose you had $f_{X,Y}(x,y) = begin{cases}1/4,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$
How would you find a CDF for $Z=XY$?
I know it's of the form
$$F_Z(z) = P(XY leq z) = iint_{{(x,y):xyleq z}}f_{X,Y}(x,y)~dx~dy$$
I'm just a bit unclear on how to set up the bounds of the integral.
calculus probability integration
edited Nov 23 '18 at 3:39
Math Lover
13.7k31435
asked Nov 22 '18 at 16:20
HumptyDumptyHumptyDumpty
33518
add a comment |
Suppose you had $f_{X,Y}(x,y) = begin{cases}1/4,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$
How would you find a CDF for $Z=XY$?
I know it's of the form
$$F_Z(z) = P(XY leq z) = iint_{{(x,y):xyleq z}}f_{X,Y}(x,y)~dx~dy$$
I'm just a bit unclear on how to set up the bounds of the integral.
calculus probability integration
edited Nov 23 '18 at 3:39
Math Lover
13.7k31435
asked Nov 22 '18 at 16:20
HumptyDumptyHumptyDumpty
33518
Draw a picture and divide the integral by x=1
– GNUSupporter 8964民主女神 地下教會
Nov 22 '18 at 16:25
add a comment |
Suppose you had $f_{X,Y}(x,y) = begin{cases}1/4,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$
How would you find a CDF for $Z=XY$?
I know it's of the form
$$F_Z(z) = P(XY leq z) = iint_{{(x,y):xyleq z}}f_{X,Y}(x,y)~dx~dy$$
I'm just a bit unclear on how to set up the bounds of the integral.
calculus probability integration
edited Nov 23 '18 at 3:39
Math Lover
13.7k31435
asked Nov 22 '18 at 16:20
HumptyDumptyHumptyDumpty
33518
Suppose you had $f_{X,Y}(x,y) = begin{cases}1/4,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$
How would you find a CDF for $Z=XY$?
I know it's of the form
$$F_Z(z) = P(XY leq z) = iint_{{(x,y):xyleq z}}f_{X,Y}(x,y)~dx~dy$$
I'm just a bit unclear on how to set up the bounds of the integral.
calculus probability integration
calculus probability integration
edited Nov 23 '18 at 3:39
Math Lover
13.7k31435
asked Nov 22 '18 at 16:20
HumptyDumptyHumptyDumpty
33518
edited Nov 23 '18 at 3:39
Math Lover
13.7k31435
asked Nov 22 '18 at 16:20
HumptyDumptyHumptyDumpty
33518
edited Nov 23 '18 at 3:39
Math Lover
13.7k31435
edited Nov 23 '18 at 3:39
Math Lover
13.7k31435
edited Nov 23 '18 at 3:39
Math Lover
13.7k31435
13.7k31435
asked Nov 22 '18 at 16:20
HumptyDumptyHumptyDumpty
33518
asked Nov 22 '18 at 16:20
HumptyDumptyHumptyDumpty
33518
asked Nov 22 '18 at 16:20
HumptyDumptyHumptyDumpty
33518
33518
Draw a picture and divide the integral by x=1
– GNUSupporter 8964民主女神 地下教會
Nov 22 '18 at 16:25
add a comment |
Draw a picture and divide the integral by x=1
– GNUSupporter 8964民主女神 地下教會
Nov 22 '18 at 16:25
Draw a picture and divide the integral by x=1
– GNUSupporter 8964民主女神 地下教會
Nov 22 '18 at 16:25
Draw a picture and divide the integral by x=1
– GNUSupporter 8964民主女神 地下教會
Nov 22 '18 at 16:25
add a comment |
2 Answers
2
active
oldest
votes
Note that $Y<X^3$ and $XY<z implies Y<z/X$; i.e., $Y < min{X^3,z/X}$.
For $X^3<z/X$, or $X < z^{1/4}$, we have
$$Y < min{X^3,z/X} = X^3;$$
otherwise
$$Y < min{X^3,z/X} = z/X.$$
Consequently,
$$Pr{XY <z} = int_{0}^{z^{1/4}}int_{0}^{x^3}frac{1}{4}dy,dx + int_{z^{1/4}}^{2}int_{0}^{z/x}frac{1}{4}dy,dx,$$
where $0 < z < 16$.
answered Nov 22 '18 at 16:38
Math LoverMath Lover
13.7k31435
I don't think I understand your use of min. What is min telling us about the limits?
– HumptyDumpty
Nov 22 '18 at 16:50
@HumptyDumpty Note that $Y < X^3$ and $Y < z/X$. For $X < z^{1/4}$, $X^3 < z/X$. Therefore, $Y < X^3< z/X$. So $0 < Y < X^3$ dictates the integration limits. Similar arguments can be used regarding other integration limits.
– Math Lover
Nov 22 '18 at 17:35
@HumptyDumpty If $Y< X^3$ and $Y<z/X$ then $Y<min{X^3,z/X}$. So $$mathsf P(XY<z)=int_0^2int_0^{min{x^3, z/x}} tfrac 14~mathsf dy~mathsf dx$$
– Graham Kemp
Nov 22 '18 at 22:54
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{0}^{2}{1 over 4}
{bracks{0 < z/x < x^{3}} over verts{x}},dd x & =
{1 over 4}bracks{z > 0}
int_{0}^{2}{bracks{x > z^{1/4}} over x},dd x
\[5mm] & =
{1 over 4}bracks{z > 0}bracks{z^{1/4} < 2}int_{z^{1/4}}^{2}
{dd xover x}
\[5mm] & =
{1 over 4}bracks{0 < z < 16}lnpars{2 over z^{1/4}}
end{align}
answered Nov 22 '18 at 20:23
Felix MarinFelix Marin
67.2k7107141
add a comment |
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2 Answers
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2 Answers
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Note that $Y<X^3$ and $XY<z implies Y<z/X$; i.e., $Y < min{X^3,z/X}$.
For $X^3<z/X$, or $X < z^{1/4}$, we have
$$Y < min{X^3,z/X} = X^3;$$
otherwise
$$Y < min{X^3,z/X} = z/X.$$
Consequently,
$$Pr{XY <z} = int_{0}^{z^{1/4}}int_{0}^{x^3}frac{1}{4}dy,dx + int_{z^{1/4}}^{2}int_{0}^{z/x}frac{1}{4}dy,dx,$$
where $0 < z < 16$.
answered Nov 22 '18 at 16:38
Math LoverMath Lover
13.7k31435
I don't think I understand your use of min. What is min telling us about the limits?
– HumptyDumpty
Nov 22 '18 at 16:50
@HumptyDumpty Note that $Y < X^3$ and $Y < z/X$. For $X < z^{1/4}$, $X^3 < z/X$. Therefore, $Y < X^3< z/X$. So $0 < Y < X^3$ dictates the integration limits. Similar arguments can be used regarding other integration limits.
– Math Lover
Nov 22 '18 at 17:35
@HumptyDumpty If $Y< X^3$ and $Y<z/X$ then $Y<min{X^3,z/X}$. So $$mathsf P(XY<z)=int_0^2int_0^{min{x^3, z/x}} tfrac 14~mathsf dy~mathsf dx$$
– Graham Kemp
Nov 22 '18 at 22:54
add a comment |
Note that $Y<X^3$ and $XY<z implies Y<z/X$; i.e., $Y < min{X^3,z/X}$.
For $X^3<z/X$, or $X < z^{1/4}$, we have
$$Y < min{X^3,z/X} = X^3;$$
otherwise
$$Y < min{X^3,z/X} = z/X.$$
Consequently,
$$Pr{XY <z} = int_{0}^{z^{1/4}}int_{0}^{x^3}frac{1}{4}dy,dx + int_{z^{1/4}}^{2}int_{0}^{z/x}frac{1}{4}dy,dx,$$
where $0 < z < 16$.
answered Nov 22 '18 at 16:38
Math LoverMath Lover
13.7k31435
I don't think I understand your use of min. What is min telling us about the limits?
– HumptyDumpty
Nov 22 '18 at 16:50
@HumptyDumpty Note that $Y < X^3$ and $Y < z/X$. For $X < z^{1/4}$, $X^3 < z/X$. Therefore, $Y < X^3< z/X$. So $0 < Y < X^3$ dictates the integration limits. Similar arguments can be used regarding other integration limits.
– Math Lover
Nov 22 '18 at 17:35
@HumptyDumpty If $Y< X^3$ and $Y<z/X$ then $Y<min{X^3,z/X}$. So $$mathsf P(XY<z)=int_0^2int_0^{min{x^3, z/x}} tfrac 14~mathsf dy~mathsf dx$$
– Graham Kemp
Nov 22 '18 at 22:54
add a comment |
Note that $Y<X^3$ and $XY<z implies Y<z/X$; i.e., $Y < min{X^3,z/X}$.
For $X^3<z/X$, or $X < z^{1/4}$, we have
$$Y < min{X^3,z/X} = X^3;$$
otherwise
$$Y < min{X^3,z/X} = z/X.$$
Consequently,
$$Pr{XY <z} = int_{0}^{z^{1/4}}int_{0}^{x^3}frac{1}{4}dy,dx + int_{z^{1/4}}^{2}int_{0}^{z/x}frac{1}{4}dy,dx,$$
where $0 < z < 16$.
answered Nov 22 '18 at 16:38
Math LoverMath Lover
13.7k31435
Note that $Y<X^3$ and $XY<z implies Y<z/X$; i.e., $Y < min{X^3,z/X}$.
For $X^3<z/X$, or $X < z^{1/4}$, we have
$$Y < min{X^3,z/X} = X^3;$$
otherwise
$$Y < min{X^3,z/X} = z/X.$$
Consequently,
$$Pr{XY <z} = int_{0}^{z^{1/4}}int_{0}^{x^3}frac{1}{4}dy,dx + int_{z^{1/4}}^{2}int_{0}^{z/x}frac{1}{4}dy,dx,$$
where $0 < z < 16$.
answered Nov 22 '18 at 16:38
Math LoverMath Lover
13.7k31435
answered Nov 22 '18 at 16:38
Math LoverMath Lover
13.7k31435
answered Nov 22 '18 at 16:38
Math LoverMath Lover
13.7k31435
answered Nov 22 '18 at 16:38
Math LoverMath Lover
13.7k31435
13.7k31435
I don't think I understand your use of min. What is min telling us about the limits?
– HumptyDumpty
Nov 22 '18 at 16:50
@HumptyDumpty Note that $Y < X^3$ and $Y < z/X$. For $X < z^{1/4}$, $X^3 < z/X$. Therefore, $Y < X^3< z/X$. So $0 < Y < X^3$ dictates the integration limits. Similar arguments can be used regarding other integration limits.
– Math Lover
Nov 22 '18 at 17:35
@HumptyDumpty If $Y< X^3$ and $Y<z/X$ then $Y<min{X^3,z/X}$. So $$mathsf P(XY<z)=int_0^2int_0^{min{x^3, z/x}} tfrac 14~mathsf dy~mathsf dx$$
– Graham Kemp
Nov 22 '18 at 22:54
add a comment |
I don't think I understand your use of min. What is min telling us about the limits?
– HumptyDumpty
Nov 22 '18 at 16:50
@HumptyDumpty Note that $Y < X^3$ and $Y < z/X$. For $X < z^{1/4}$, $X^3 < z/X$. Therefore, $Y < X^3< z/X$. So $0 < Y < X^3$ dictates the integration limits. Similar arguments can be used regarding other integration limits.
– Math Lover
Nov 22 '18 at 17:35
@HumptyDumpty If $Y< X^3$ and $Y<z/X$ then $Y<min{X^3,z/X}$. So $$mathsf P(XY<z)=int_0^2int_0^{min{x^3, z/x}} tfrac 14~mathsf dy~mathsf dx$$
– Graham Kemp
Nov 22 '18 at 22:54
I don't think I understand your use of min. What is min telling us about the limits?
– HumptyDumpty
Nov 22 '18 at 16:50
I don't think I understand your use of min. What is min telling us about the limits?
– HumptyDumpty
Nov 22 '18 at 16:50
@HumptyDumpty Note that $Y < X^3$ and $Y < z/X$. For $X < z^{1/4}$, $X^3 < z/X$. Therefore, $Y < X^3< z/X$. So $0 < Y < X^3$ dictates the integration limits. Similar arguments can be used regarding other integration limits.
– Math Lover
Nov 22 '18 at 17:35
@HumptyDumpty Note that $Y < X^3$ and $Y < z/X$. For $X < z^{1/4}$, $X^3 < z/X$. Therefore, $Y < X^3< z/X$. So $0 < Y < X^3$ dictates the integration limits. Similar arguments can be used regarding other integration limits.
– Math Lover
Nov 22 '18 at 17:35
@HumptyDumpty If $Y< X^3$ and $Y<z/X$ then $Y<min{X^3,z/X}$. So $$mathsf P(XY<z)=int_0^2int_0^{min{x^3, z/x}} tfrac 14~mathsf dy~mathsf dx$$
– Graham Kemp
Nov 22 '18 at 22:54
@HumptyDumpty If $Y< X^3$ and $Y<z/X$ then $Y<min{X^3,z/X}$. So $$mathsf P(XY<z)=int_0^2int_0^{min{x^3, z/x}} tfrac 14~mathsf dy~mathsf dx$$
– Graham Kemp
Nov 22 '18 at 22:54
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{0}^{2}{1 over 4}
{bracks{0 < z/x < x^{3}} over verts{x}},dd x & =
{1 over 4}bracks{z > 0}
int_{0}^{2}{bracks{x > z^{1/4}} over x},dd x
\[5mm] & =
{1 over 4}bracks{z > 0}bracks{z^{1/4} < 2}int_{z^{1/4}}^{2}
{dd xover x}
\[5mm] & =
{1 over 4}bracks{0 < z < 16}lnpars{2 over z^{1/4}}
end{align}
answered Nov 22 '18 at 20:23
Felix MarinFelix Marin
67.2k7107141
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{0}^{2}{1 over 4}
{bracks{0 < z/x < x^{3}} over verts{x}},dd x & =
{1 over 4}bracks{z > 0}
int_{0}^{2}{bracks{x > z^{1/4}} over x},dd x
\[5mm] & =
{1 over 4}bracks{z > 0}bracks{z^{1/4} < 2}int_{z^{1/4}}^{2}
{dd xover x}
\[5mm] & =
{1 over 4}bracks{0 < z < 16}lnpars{2 over z^{1/4}}
end{align}
answered Nov 22 '18 at 20:23
Felix MarinFelix Marin
67.2k7107141
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{0}^{2}{1 over 4}
{bracks{0 < z/x < x^{3}} over verts{x}},dd x & =
{1 over 4}bracks{z > 0}
int_{0}^{2}{bracks{x > z^{1/4}} over x},dd x
\[5mm] & =
{1 over 4}bracks{z > 0}bracks{z^{1/4} < 2}int_{z^{1/4}}^{2}
{dd xover x}
\[5mm] & =
{1 over 4}bracks{0 < z < 16}lnpars{2 over z^{1/4}}
end{align}
answered Nov 22 '18 at 20:23
Felix MarinFelix Marin
67.2k7107141
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{0}^{2}{1 over 4}
{bracks{0 < z/x < x^{3}} over verts{x}},dd x & =
{1 over 4}bracks{z > 0}
int_{0}^{2}{bracks{x > z^{1/4}} over x},dd x
\[5mm] & =
{1 over 4}bracks{z > 0}bracks{z^{1/4} < 2}int_{z^{1/4}}^{2}
{dd xover x}
\[5mm] & =
{1 over 4}bracks{0 < z < 16}lnpars{2 over z^{1/4}}
end{align}
answered Nov 22 '18 at 20:23
Felix MarinFelix Marin
67.2k7107141
answered Nov 22 '18 at 20:23
Felix MarinFelix Marin
67.2k7107141
answered Nov 22 '18 at 20:23
Felix MarinFelix Marin
67.2k7107141
answered Nov 22 '18 at 20:23
Felix MarinFelix Marin
67.2k7107141
67.2k7107141
add a comment |
add a comment |
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Draw a picture and divide the integral by x=1
– GNUSupporter 8964民主女神 地下教會
Nov 22 '18 at 16:25