PDF of Z = XY for Jointly Uniform (X,Y) with Parabolic Region
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"Suppose that $(X,Y)$ is uniformly distributed on the subset of $; mathbb R^2$ defined by the inequalities $0 < X < 1$ and $0 < Y < X^2$.
Determine the probability density function of the random variable $Z = XY$"
The $X^2$ in the $Y$ interval is seriously throwing me off and I'm not sure where to start. I have
$$f_{X,Y}(x,y) = 3 qquad (x,y) epsilon;(0,1)times(0,X^2)$$
$$f_{X,Y}(x,y) = 0 qquad qquad qquad quad otherwise$$
But I do not know where to go from here. I need to be able to explain my answer so I'd like to understand how/why it works. Thanks!
probability transformation independence uniform-distribution
asked Nov 21 at 21:12
Angus Pointer
186
add a comment |
up vote
1
down vote
favorite
"Suppose that $(X,Y)$ is uniformly distributed on the subset of $; mathbb R^2$ defined by the inequalities $0 < X < 1$ and $0 < Y < X^2$.
Determine the probability density function of the random variable $Z = XY$"
The $X^2$ in the $Y$ interval is seriously throwing me off and I'm not sure where to start. I have
$$f_{X,Y}(x,y) = 3 qquad (x,y) epsilon;(0,1)times(0,X^2)$$
$$f_{X,Y}(x,y) = 0 qquad qquad qquad quad otherwise$$
But I do not know where to go from here. I need to be able to explain my answer so I'd like to understand how/why it works. Thanks!
probability transformation independence uniform-distribution
asked Nov 21 at 21:12
Angus Pointer
186
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
"Suppose that $(X,Y)$ is uniformly distributed on the subset of $; mathbb R^2$ defined by the inequalities $0 < X < 1$ and $0 < Y < X^2$.
Determine the probability density function of the random variable $Z = XY$"
The $X^2$ in the $Y$ interval is seriously throwing me off and I'm not sure where to start. I have
$$f_{X,Y}(x,y) = 3 qquad (x,y) epsilon;(0,1)times(0,X^2)$$
$$f_{X,Y}(x,y) = 0 qquad qquad qquad quad otherwise$$
But I do not know where to go from here. I need to be able to explain my answer so I'd like to understand how/why it works. Thanks!
probability transformation independence uniform-distribution
asked Nov 21 at 21:12
Angus Pointer
186
"Suppose that $(X,Y)$ is uniformly distributed on the subset of $; mathbb R^2$ defined by the inequalities $0 < X < 1$ and $0 < Y < X^2$.
Determine the probability density function of the random variable $Z = XY$"
The $X^2$ in the $Y$ interval is seriously throwing me off and I'm not sure where to start. I have
$$f_{X,Y}(x,y) = 3 qquad (x,y) epsilon;(0,1)times(0,X^2)$$
$$f_{X,Y}(x,y) = 0 qquad qquad qquad quad otherwise$$
But I do not know where to go from here. I need to be able to explain my answer so I'd like to understand how/why it works. Thanks!
probability transformation independence uniform-distribution
probability transformation independence uniform-distribution
asked Nov 21 at 21:12
Angus Pointer
186
asked Nov 21 at 21:12
Angus Pointer
186
edited Nov 21 at 21:33
asked Nov 21 at 21:12
Angus Pointer
186
asked Nov 21 at 21:12
Angus Pointer
186
asked Nov 21 at 21:12
Angus Pointer
186
186
add a comment |
add a comment |
2 Answers
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oldest
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up vote
1
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accepted
You have found that$f_{X,Y}(x,y)=3cdotmathbf 1_{0<x<1, 0<y<x^2}$ . $checkmark$
From there use the Jacobian Transformation Theorem.
$$begin{align}f_{X,Z}(x,z) &= begin{Vmatrix}dfrac{partial[x,z/x]}{partial[x,z]}end{Vmatrix}~f_{X,Y}(x,z/x)\[1ex] &= dfrac 1{lvert xrvert}cdot 3cdotmathbf 1_{0< x< 1~,~ 0< z/x< x^2}\[1ex] &= dfrac 3{x}cdotmathbf 1_{0< z< 1~,~ sqrt[3]z< x< 1}end{align}$$
After which integration with respect to $x$ finds the marginal distribution. (Notice the support)
$$begin{align}f_Z(z) &= int_Bbb R f_{X,Z}(x,z)~mathsf d x\[1ex] &= int_{sqrt[3]z}^1frac 3x~mathsf d xcdotmathbf 1_{0<z< 1}\[1ex] &= -3ln(sqrt[3]z)cdotmathbf 1_{0< z< 1}\[1ex] &= -ln(z)cdotmathbf 1_{0< z< 1}end{align}$$
answered Nov 22 at 3:09
Graham Kemp
84.6k43378
Thanks! Could you please explain what the 1 (0 < z < 1) means?
– Angus Pointer
Nov 22 at 15:46
1
It is an Indicator random variable, which is equal to 1 when the index is true and zero otherwise.
– Graham Kemp
Nov 22 at 20:04
add a comment |
up vote
1
down vote
I've looked over some examples and believe I have a solution.
$$f_{X,Y}(x,y) =
begin{cases}
3 &&& text{for} quad0<X<1, & 0<Y<X^2 \
0 &&& text{otherwise}
end{cases}
$$
Then,
$$F_Z(z) = begin{cases}
0 &&& Z leq 0 \
Bbb P (Z leq z) = Bbb P (XY leq z) = Bbb P(Y leq frac{1}{X}z) &&& 0<Z<X^2 \
1 &&& Z geq X^2
end{cases}
$$
Our support region is $$A = {(X,Y); |; 0 < X < 1,; ; 0 < Y < X^2} cap {(X,Y) ;| ;Y leq frac{1}{X} z}$$
Integrating $f_{X,Y}(x,y)$ twice over this region, where the two "curves" $Y=X^2$ and $Y = frac{1}{X}z$ intersect at $(z^{frac{1}{3}},z^{frac{2}{3}})$
$$ Bbb P(Y leq frac{1}{X}z) = iint_A3 dx dy = 3intlimits^{z^{frac{1}{3}}}_0intlimits^{x^{2}}_0 dy dx + 3intlimits^{1}_{z^{frac{1}{3}}}intlimits^{frac{z}{x}}_{0} dy dx \
= x^3 biggrvert_{0}^{z^{frac{1}{3}}}+3z ln{x}biggrvert^0_{z^{frac{1}{3}}} = z - 3zln{z^{frac{1}{3}}} = z(1-ln{z})$$
So we have;
$$F_Z(z) = begin{cases} 0 & Zleq 0 \
z(1-ln{z}) & 0 < Z < X^2 \
1 & Zgeq X^2
end{cases}$$
then differentiate to obtain $f_X(x)$;
$$f_X(x) = begin{cases}
-ln{z} & 0 < Z < X^2 \
0 & text{otherwise}
end{cases}$$
I'm still unsure about my interval for Z, but otherwise I believe this is correct?
answered Nov 22 at 0:36
Angus Pointer
186
1
$Z$ will be supported over $(0,1)$, [the bounds should not mention either $X$ or $Y$, but be the result of their bounds, and we have $0< XY< 1cdot1^2$], and of course you are looking for $f_Z(z)$ when you differentiate, rather than $f_X(x)$.
– Graham Kemp
Nov 22 at 3:12
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have found that$f_{X,Y}(x,y)=3cdotmathbf 1_{0<x<1, 0<y<x^2}$ . $checkmark$
From there use the Jacobian Transformation Theorem.
$$begin{align}f_{X,Z}(x,z) &= begin{Vmatrix}dfrac{partial[x,z/x]}{partial[x,z]}end{Vmatrix}~f_{X,Y}(x,z/x)\[1ex] &= dfrac 1{lvert xrvert}cdot 3cdotmathbf 1_{0< x< 1~,~ 0< z/x< x^2}\[1ex] &= dfrac 3{x}cdotmathbf 1_{0< z< 1~,~ sqrt[3]z< x< 1}end{align}$$
After which integration with respect to $x$ finds the marginal distribution. (Notice the support)
$$begin{align}f_Z(z) &= int_Bbb R f_{X,Z}(x,z)~mathsf d x\[1ex] &= int_{sqrt[3]z}^1frac 3x~mathsf d xcdotmathbf 1_{0<z< 1}\[1ex] &= -3ln(sqrt[3]z)cdotmathbf 1_{0< z< 1}\[1ex] &= -ln(z)cdotmathbf 1_{0< z< 1}end{align}$$
answered Nov 22 at 3:09
Graham Kemp
84.6k43378
Thanks! Could you please explain what the 1 (0 < z < 1) means?
– Angus Pointer
Nov 22 at 15:46
1
It is an Indicator random variable, which is equal to 1 when the index is true and zero otherwise.
– Graham Kemp
Nov 22 at 20:04
add a comment |
up vote
1
down vote
accepted
You have found that$f_{X,Y}(x,y)=3cdotmathbf 1_{0<x<1, 0<y<x^2}$ . $checkmark$
From there use the Jacobian Transformation Theorem.
$$begin{align}f_{X,Z}(x,z) &= begin{Vmatrix}dfrac{partial[x,z/x]}{partial[x,z]}end{Vmatrix}~f_{X,Y}(x,z/x)\[1ex] &= dfrac 1{lvert xrvert}cdot 3cdotmathbf 1_{0< x< 1~,~ 0< z/x< x^2}\[1ex] &= dfrac 3{x}cdotmathbf 1_{0< z< 1~,~ sqrt[3]z< x< 1}end{align}$$
After which integration with respect to $x$ finds the marginal distribution. (Notice the support)
$$begin{align}f_Z(z) &= int_Bbb R f_{X,Z}(x,z)~mathsf d x\[1ex] &= int_{sqrt[3]z}^1frac 3x~mathsf d xcdotmathbf 1_{0<z< 1}\[1ex] &= -3ln(sqrt[3]z)cdotmathbf 1_{0< z< 1}\[1ex] &= -ln(z)cdotmathbf 1_{0< z< 1}end{align}$$
answered Nov 22 at 3:09
Graham Kemp
84.6k43378
Thanks! Could you please explain what the 1 (0 < z < 1) means?
– Angus Pointer
Nov 22 at 15:46
1
It is an Indicator random variable, which is equal to 1 when the index is true and zero otherwise.
– Graham Kemp
Nov 22 at 20:04
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have found that$f_{X,Y}(x,y)=3cdotmathbf 1_{0<x<1, 0<y<x^2}$ . $checkmark$
From there use the Jacobian Transformation Theorem.
$$begin{align}f_{X,Z}(x,z) &= begin{Vmatrix}dfrac{partial[x,z/x]}{partial[x,z]}end{Vmatrix}~f_{X,Y}(x,z/x)\[1ex] &= dfrac 1{lvert xrvert}cdot 3cdotmathbf 1_{0< x< 1~,~ 0< z/x< x^2}\[1ex] &= dfrac 3{x}cdotmathbf 1_{0< z< 1~,~ sqrt[3]z< x< 1}end{align}$$
After which integration with respect to $x$ finds the marginal distribution. (Notice the support)
$$begin{align}f_Z(z) &= int_Bbb R f_{X,Z}(x,z)~mathsf d x\[1ex] &= int_{sqrt[3]z}^1frac 3x~mathsf d xcdotmathbf 1_{0<z< 1}\[1ex] &= -3ln(sqrt[3]z)cdotmathbf 1_{0< z< 1}\[1ex] &= -ln(z)cdotmathbf 1_{0< z< 1}end{align}$$
answered Nov 22 at 3:09
Graham Kemp
84.6k43378
You have found that$f_{X,Y}(x,y)=3cdotmathbf 1_{0<x<1, 0<y<x^2}$ . $checkmark$
From there use the Jacobian Transformation Theorem.
$$begin{align}f_{X,Z}(x,z) &= begin{Vmatrix}dfrac{partial[x,z/x]}{partial[x,z]}end{Vmatrix}~f_{X,Y}(x,z/x)\[1ex] &= dfrac 1{lvert xrvert}cdot 3cdotmathbf 1_{0< x< 1~,~ 0< z/x< x^2}\[1ex] &= dfrac 3{x}cdotmathbf 1_{0< z< 1~,~ sqrt[3]z< x< 1}end{align}$$
After which integration with respect to $x$ finds the marginal distribution. (Notice the support)
$$begin{align}f_Z(z) &= int_Bbb R f_{X,Z}(x,z)~mathsf d x\[1ex] &= int_{sqrt[3]z}^1frac 3x~mathsf d xcdotmathbf 1_{0<z< 1}\[1ex] &= -3ln(sqrt[3]z)cdotmathbf 1_{0< z< 1}\[1ex] &= -ln(z)cdotmathbf 1_{0< z< 1}end{align}$$
answered Nov 22 at 3:09
Graham Kemp
84.6k43378
answered Nov 22 at 3:09
Graham Kemp
84.6k43378
answered Nov 22 at 3:09
Graham Kemp
84.6k43378
answered Nov 22 at 3:09
Graham Kemp
84.6k43378
84.6k43378
Thanks! Could you please explain what the 1 (0 < z < 1) means?
– Angus Pointer
Nov 22 at 15:46
1
It is an Indicator random variable, which is equal to 1 when the index is true and zero otherwise.
– Graham Kemp
Nov 22 at 20:04
add a comment |
Thanks! Could you please explain what the 1 (0 < z < 1) means?
– Angus Pointer
Nov 22 at 15:46
1
It is an Indicator random variable, which is equal to 1 when the index is true and zero otherwise.
– Graham Kemp
Nov 22 at 20:04
Thanks! Could you please explain what the 1 (0 < z < 1) means?
– Angus Pointer
Nov 22 at 15:46
Thanks! Could you please explain what the 1 (0 < z < 1) means?
– Angus Pointer
Nov 22 at 15:46
1
1
It is an Indicator random variable, which is equal to 1 when the index is true and zero otherwise.
– Graham Kemp
Nov 22 at 20:04
It is an Indicator random variable, which is equal to 1 when the index is true and zero otherwise.
– Graham Kemp
Nov 22 at 20:04
add a comment |
up vote
1
down vote
I've looked over some examples and believe I have a solution.
$$f_{X,Y}(x,y) =
begin{cases}
3 &&& text{for} quad0<X<1, & 0<Y<X^2 \
0 &&& text{otherwise}
end{cases}
$$
Then,
$$F_Z(z) = begin{cases}
0 &&& Z leq 0 \
Bbb P (Z leq z) = Bbb P (XY leq z) = Bbb P(Y leq frac{1}{X}z) &&& 0<Z<X^2 \
1 &&& Z geq X^2
end{cases}
$$
Our support region is $$A = {(X,Y); |; 0 < X < 1,; ; 0 < Y < X^2} cap {(X,Y) ;| ;Y leq frac{1}{X} z}$$
Integrating $f_{X,Y}(x,y)$ twice over this region, where the two "curves" $Y=X^2$ and $Y = frac{1}{X}z$ intersect at $(z^{frac{1}{3}},z^{frac{2}{3}})$
$$ Bbb P(Y leq frac{1}{X}z) = iint_A3 dx dy = 3intlimits^{z^{frac{1}{3}}}_0intlimits^{x^{2}}_0 dy dx + 3intlimits^{1}_{z^{frac{1}{3}}}intlimits^{frac{z}{x}}_{0} dy dx \
= x^3 biggrvert_{0}^{z^{frac{1}{3}}}+3z ln{x}biggrvert^0_{z^{frac{1}{3}}} = z - 3zln{z^{frac{1}{3}}} = z(1-ln{z})$$
So we have;
$$F_Z(z) = begin{cases} 0 & Zleq 0 \
z(1-ln{z}) & 0 < Z < X^2 \
1 & Zgeq X^2
end{cases}$$
then differentiate to obtain $f_X(x)$;
$$f_X(x) = begin{cases}
-ln{z} & 0 < Z < X^2 \
0 & text{otherwise}
end{cases}$$
I'm still unsure about my interval for Z, but otherwise I believe this is correct?
answered Nov 22 at 0:36
Angus Pointer
186
1
$Z$ will be supported over $(0,1)$, [the bounds should not mention either $X$ or $Y$, but be the result of their bounds, and we have $0< XY< 1cdot1^2$], and of course you are looking for $f_Z(z)$ when you differentiate, rather than $f_X(x)$.
– Graham Kemp
Nov 22 at 3:12
add a comment |
up vote
1
down vote
I've looked over some examples and believe I have a solution.
$$f_{X,Y}(x,y) =
begin{cases}
3 &&& text{for} quad0<X<1, & 0<Y<X^2 \
0 &&& text{otherwise}
end{cases}
$$
Then,
$$F_Z(z) = begin{cases}
0 &&& Z leq 0 \
Bbb P (Z leq z) = Bbb P (XY leq z) = Bbb P(Y leq frac{1}{X}z) &&& 0<Z<X^2 \
1 &&& Z geq X^2
end{cases}
$$
Our support region is $$A = {(X,Y); |; 0 < X < 1,; ; 0 < Y < X^2} cap {(X,Y) ;| ;Y leq frac{1}{X} z}$$
Integrating $f_{X,Y}(x,y)$ twice over this region, where the two "curves" $Y=X^2$ and $Y = frac{1}{X}z$ intersect at $(z^{frac{1}{3}},z^{frac{2}{3}})$
$$ Bbb P(Y leq frac{1}{X}z) = iint_A3 dx dy = 3intlimits^{z^{frac{1}{3}}}_0intlimits^{x^{2}}_0 dy dx + 3intlimits^{1}_{z^{frac{1}{3}}}intlimits^{frac{z}{x}}_{0} dy dx \
= x^3 biggrvert_{0}^{z^{frac{1}{3}}}+3z ln{x}biggrvert^0_{z^{frac{1}{3}}} = z - 3zln{z^{frac{1}{3}}} = z(1-ln{z})$$
So we have;
$$F_Z(z) = begin{cases} 0 & Zleq 0 \
z(1-ln{z}) & 0 < Z < X^2 \
1 & Zgeq X^2
end{cases}$$
then differentiate to obtain $f_X(x)$;
$$f_X(x) = begin{cases}
-ln{z} & 0 < Z < X^2 \
0 & text{otherwise}
end{cases}$$
I'm still unsure about my interval for Z, but otherwise I believe this is correct?
answered Nov 22 at 0:36
Angus Pointer
186
1
$Z$ will be supported over $(0,1)$, [the bounds should not mention either $X$ or $Y$, but be the result of their bounds, and we have $0< XY< 1cdot1^2$], and of course you are looking for $f_Z(z)$ when you differentiate, rather than $f_X(x)$.
– Graham Kemp
Nov 22 at 3:12
add a comment |
up vote
1
down vote
up vote
1
down vote
I've looked over some examples and believe I have a solution.
$$f_{X,Y}(x,y) =
begin{cases}
3 &&& text{for} quad0<X<1, & 0<Y<X^2 \
0 &&& text{otherwise}
end{cases}
$$
Then,
$$F_Z(z) = begin{cases}
0 &&& Z leq 0 \
Bbb P (Z leq z) = Bbb P (XY leq z) = Bbb P(Y leq frac{1}{X}z) &&& 0<Z<X^2 \
1 &&& Z geq X^2
end{cases}
$$
Our support region is $$A = {(X,Y); |; 0 < X < 1,; ; 0 < Y < X^2} cap {(X,Y) ;| ;Y leq frac{1}{X} z}$$
Integrating $f_{X,Y}(x,y)$ twice over this region, where the two "curves" $Y=X^2$ and $Y = frac{1}{X}z$ intersect at $(z^{frac{1}{3}},z^{frac{2}{3}})$
$$ Bbb P(Y leq frac{1}{X}z) = iint_A3 dx dy = 3intlimits^{z^{frac{1}{3}}}_0intlimits^{x^{2}}_0 dy dx + 3intlimits^{1}_{z^{frac{1}{3}}}intlimits^{frac{z}{x}}_{0} dy dx \
= x^3 biggrvert_{0}^{z^{frac{1}{3}}}+3z ln{x}biggrvert^0_{z^{frac{1}{3}}} = z - 3zln{z^{frac{1}{3}}} = z(1-ln{z})$$
So we have;
$$F_Z(z) = begin{cases} 0 & Zleq 0 \
z(1-ln{z}) & 0 < Z < X^2 \
1 & Zgeq X^2
end{cases}$$
then differentiate to obtain $f_X(x)$;
$$f_X(x) = begin{cases}
-ln{z} & 0 < Z < X^2 \
0 & text{otherwise}
end{cases}$$
I'm still unsure about my interval for Z, but otherwise I believe this is correct?
answered Nov 22 at 0:36
Angus Pointer
186
I've looked over some examples and believe I have a solution.
$$f_{X,Y}(x,y) =
begin{cases}
3 &&& text{for} quad0<X<1, & 0<Y<X^2 \
0 &&& text{otherwise}
end{cases}
$$
Then,
$$F_Z(z) = begin{cases}
0 &&& Z leq 0 \
Bbb P (Z leq z) = Bbb P (XY leq z) = Bbb P(Y leq frac{1}{X}z) &&& 0<Z<X^2 \
1 &&& Z geq X^2
end{cases}
$$
Our support region is $$A = {(X,Y); |; 0 < X < 1,; ; 0 < Y < X^2} cap {(X,Y) ;| ;Y leq frac{1}{X} z}$$
Integrating $f_{X,Y}(x,y)$ twice over this region, where the two "curves" $Y=X^2$ and $Y = frac{1}{X}z$ intersect at $(z^{frac{1}{3}},z^{frac{2}{3}})$
$$ Bbb P(Y leq frac{1}{X}z) = iint_A3 dx dy = 3intlimits^{z^{frac{1}{3}}}_0intlimits^{x^{2}}_0 dy dx + 3intlimits^{1}_{z^{frac{1}{3}}}intlimits^{frac{z}{x}}_{0} dy dx \
= x^3 biggrvert_{0}^{z^{frac{1}{3}}}+3z ln{x}biggrvert^0_{z^{frac{1}{3}}} = z - 3zln{z^{frac{1}{3}}} = z(1-ln{z})$$
So we have;
$$F_Z(z) = begin{cases} 0 & Zleq 0 \
z(1-ln{z}) & 0 < Z < X^2 \
1 & Zgeq X^2
end{cases}$$
then differentiate to obtain $f_X(x)$;
$$f_X(x) = begin{cases}
-ln{z} & 0 < Z < X^2 \
0 & text{otherwise}
end{cases}$$
I'm still unsure about my interval for Z, but otherwise I believe this is correct?
answered Nov 22 at 0:36
Angus Pointer
186
answered Nov 22 at 0:36
Angus Pointer
186
answered Nov 22 at 0:36
Angus Pointer
186
answered Nov 22 at 0:36
Angus Pointer
186
186
1
$Z$ will be supported over $(0,1)$, [the bounds should not mention either $X$ or $Y$, but be the result of their bounds, and we have $0< XY< 1cdot1^2$], and of course you are looking for $f_Z(z)$ when you differentiate, rather than $f_X(x)$.
– Graham Kemp
Nov 22 at 3:12
add a comment |
1
$Z$ will be supported over $(0,1)$, [the bounds should not mention either $X$ or $Y$, but be the result of their bounds, and we have $0< XY< 1cdot1^2$], and of course you are looking for $f_Z(z)$ when you differentiate, rather than $f_X(x)$.
– Graham Kemp
Nov 22 at 3:12
1
1
$Z$ will be supported over $(0,1)$, [the bounds should not mention either $X$ or $Y$, but be the result of their bounds, and we have $0< XY< 1cdot1^2$], and of course you are looking for $f_Z(z)$ when you differentiate, rather than $f_X(x)$.
– Graham Kemp
Nov 22 at 3:12
$Z$ will be supported over $(0,1)$, [the bounds should not mention either $X$ or $Y$, but be the result of their bounds, and we have $0< XY< 1cdot1^2$], and of course you are looking for $f_Z(z)$ when you differentiate, rather than $f_X(x)$.
– Graham Kemp
Nov 22 at 3:12
add a comment |
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