Mixing
Finding the probability density of the sum $Z=X+Y$ given a joint density function.
$begingroup$
I'm given the joint density $$f_{X,Y}(x,y)=
left{begin{matrix}e^{-y}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
and I'm told to the find the probability density of the sum $Z = X+Y$.
The domains are what are throwing me off. Any comments on the domain or other incorrect parts of the solution are appreciated! Thanks.
My solution:
$$F_{Z}(z) = P(Z leq z) = P(X+Y leq z) = intint_{x+y leq z}f_{X,Y}(x,y)dxdy$$
$$=int_{-infty}^{infty}dxint_{-infty}^{z-x}f_{X,Y}(x,y)dxdy.$$
$$ $$
$$f_{Z}(z) = frac{d}{dz}F_{Z}(z)=int_{-infty}^{infty}f_{X,Y}(x,z-x)dx.$$
$$ $$
NOTE: $0leq x leq y le infty$. So, $x geq 0$ and $x leq frac{z}{2}$.
So, $$f_{Z}(z) =int_{0}^{z/2}f_{X,Y}(x,z-x)dx $$
$$= int_{0}^{z/2}e^{x-z}dx$$
$$= e^{-z}int_{0}^{z/2}e^{x}dx = e^{-z}left [ e^{x} right]Big|_0^frac{z}{2} $$
$$ = e^{-y}[e^{frac{z}{2}}-1]. $$ $$
Thus,
$$f_{Z}(z)=
left{begin{matrix}e^{-z}[e^{frac{z}{2}}-1], mbox{ } z geq 0
\ 0, mbox{ z < 0}end{matrix}right.$$
probability random-variables density-function
asked Jan 23 at 21:14
beepboopbeepboopbeepboopbeepboop
10810
$endgroup$
add a comment |
$begingroup$
I'm given the joint density $$f_{X,Y}(x,y)=
left{begin{matrix}e^{-y}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
and I'm told to the find the probability density of the sum $Z = X+Y$.
The domains are what are throwing me off. Any comments on the domain or other incorrect parts of the solution are appreciated! Thanks.
My solution:
$$F_{Z}(z) = P(Z leq z) = P(X+Y leq z) = intint_{x+y leq z}f_{X,Y}(x,y)dxdy$$
$$=int_{-infty}^{infty}dxint_{-infty}^{z-x}f_{X,Y}(x,y)dxdy.$$
$$ $$
$$f_{Z}(z) = frac{d}{dz}F_{Z}(z)=int_{-infty}^{infty}f_{X,Y}(x,z-x)dx.$$
$$ $$
NOTE: $0leq x leq y le infty$. So, $x geq 0$ and $x leq frac{z}{2}$.
So, $$f_{Z}(z) =int_{0}^{z/2}f_{X,Y}(x,z-x)dx $$
$$= int_{0}^{z/2}e^{x-z}dx$$
$$= e^{-z}int_{0}^{z/2}e^{x}dx = e^{-z}left [ e^{x} right]Big|_0^frac{z}{2} $$
$$ = e^{-y}[e^{frac{z}{2}}-1]. $$ $$
Thus,
$$f_{Z}(z)=
left{begin{matrix}e^{-z}[e^{frac{z}{2}}-1], mbox{ } z geq 0
\ 0, mbox{ z < 0}end{matrix}right.$$
probability random-variables density-function
asked Jan 23 at 21:14
beepboopbeepboopbeepboopbeepboop
10810
$endgroup$
2
$begingroup$
$f_Z(z)$ msut be a function of $z$ alone. It cannot depend on $y$.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 23:34
$begingroup$
@KaviRamaMurthy Thanks, I fixed that.
$endgroup$
– beepboopbeepboop
Jan 24 at 1:05
$begingroup$
@KaviRamaMurthy Now, I believe the answer should be correct. Would you agree?
$endgroup$
– beepboopbeepboop
Jan 24 at 1:07
$begingroup$
You were already explained, à propos one of your earlier questions on the subject, how to deal with what you call the domain, and you were given an explicit procedure. Why not try to apply it?
$endgroup$
– Did
Jan 25 at 23:39
add a comment |
$begingroup$
I'm given the joint density $$f_{X,Y}(x,y)=
left{begin{matrix}e^{-y}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
and I'm told to the find the probability density of the sum $Z = X+Y$.
The domains are what are throwing me off. Any comments on the domain or other incorrect parts of the solution are appreciated! Thanks.
My solution:
$$F_{Z}(z) = P(Z leq z) = P(X+Y leq z) = intint_{x+y leq z}f_{X,Y}(x,y)dxdy$$
$$=int_{-infty}^{infty}dxint_{-infty}^{z-x}f_{X,Y}(x,y)dxdy.$$
$$ $$
$$f_{Z}(z) = frac{d}{dz}F_{Z}(z)=int_{-infty}^{infty}f_{X,Y}(x,z-x)dx.$$
$$ $$
NOTE: $0leq x leq y le infty$. So, $x geq 0$ and $x leq frac{z}{2}$.
So, $$f_{Z}(z) =int_{0}^{z/2}f_{X,Y}(x,z-x)dx $$
$$= int_{0}^{z/2}e^{x-z}dx$$
$$= e^{-z}int_{0}^{z/2}e^{x}dx = e^{-z}left [ e^{x} right]Big|_0^frac{z}{2} $$
$$ = e^{-y}[e^{frac{z}{2}}-1]. $$ $$
Thus,
$$f_{Z}(z)=
left{begin{matrix}e^{-z}[e^{frac{z}{2}}-1], mbox{ } z geq 0
\ 0, mbox{ z < 0}end{matrix}right.$$
probability random-variables density-function
asked Jan 23 at 21:14
beepboopbeepboopbeepboopbeepboop
10810
$endgroup$
I'm given the joint density $$f_{X,Y}(x,y)=
left{begin{matrix}e^{-y}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
and I'm told to the find the probability density of the sum $Z = X+Y$.
The domains are what are throwing me off. Any comments on the domain or other incorrect parts of the solution are appreciated! Thanks.
My solution:
$$F_{Z}(z) = P(Z leq z) = P(X+Y leq z) = intint_{x+y leq z}f_{X,Y}(x,y)dxdy$$
$$=int_{-infty}^{infty}dxint_{-infty}^{z-x}f_{X,Y}(x,y)dxdy.$$
$$ $$
$$f_{Z}(z) = frac{d}{dz}F_{Z}(z)=int_{-infty}^{infty}f_{X,Y}(x,z-x)dx.$$
$$ $$
NOTE: $0leq x leq y le infty$. So, $x geq 0$ and $x leq frac{z}{2}$.
So, $$f_{Z}(z) =int_{0}^{z/2}f_{X,Y}(x,z-x)dx $$
$$= int_{0}^{z/2}e^{x-z}dx$$
$$= e^{-z}int_{0}^{z/2}e^{x}dx = e^{-z}left [ e^{x} right]Big|_0^frac{z}{2} $$
$$ = e^{-y}[e^{frac{z}{2}}-1]. $$ $$
Thus,
$$f_{Z}(z)=
left{begin{matrix}e^{-z}[e^{frac{z}{2}}-1], mbox{ } z geq 0
\ 0, mbox{ z < 0}end{matrix}right.$$
probability random-variables density-function
probability random-variables density-function
asked Jan 23 at 21:14
beepboopbeepboopbeepboopbeepboop
10810
asked Jan 23 at 21:14
beepboopbeepboopbeepboopbeepboop
10810
edited Jan 24 at 1:05
beepboopbeepboop
asked Jan 23 at 21:14
beepboopbeepboopbeepboopbeepboop
10810
asked Jan 23 at 21:14
beepboopbeepboopbeepboopbeepboop
10810
asked Jan 23 at 21:14
beepboopbeepboopbeepboopbeepboop
10810
10810
2
$begingroup$
$f_Z(z)$ msut be a function of $z$ alone. It cannot depend on $y$.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 23:34
$begingroup$
@KaviRamaMurthy Thanks, I fixed that.
$endgroup$
– beepboopbeepboop
Jan 24 at 1:05
$begingroup$
@KaviRamaMurthy Now, I believe the answer should be correct. Would you agree?
$endgroup$
– beepboopbeepboop
Jan 24 at 1:07
$begingroup$
You were already explained, à propos one of your earlier questions on the subject, how to deal with what you call the domain, and you were given an explicit procedure. Why not try to apply it?
$endgroup$
– Did
Jan 25 at 23:39
add a comment |
2
$begingroup$
$f_Z(z)$ msut be a function of $z$ alone. It cannot depend on $y$.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 23:34
$begingroup$
@KaviRamaMurthy Thanks, I fixed that.
$endgroup$
– beepboopbeepboop
Jan 24 at 1:05
$begingroup$
@KaviRamaMurthy Now, I believe the answer should be correct. Would you agree?
$endgroup$
– beepboopbeepboop
Jan 24 at 1:07
$begingroup$
You were already explained, à propos one of your earlier questions on the subject, how to deal with what you call the domain, and you were given an explicit procedure. Why not try to apply it?
$endgroup$
– Did
Jan 25 at 23:39
2
2
$begingroup$
$f_Z(z)$ msut be a function of $z$ alone. It cannot depend on $y$.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 23:34
$begingroup$
$f_Z(z)$ msut be a function of $z$ alone. It cannot depend on $y$.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 23:34
$begingroup$
@KaviRamaMurthy Thanks, I fixed that.
$endgroup$
– beepboopbeepboop
Jan 24 at 1:05
$begingroup$
@KaviRamaMurthy Thanks, I fixed that.
$endgroup$
– beepboopbeepboop
Jan 24 at 1:05
$begingroup$
@KaviRamaMurthy Now, I believe the answer should be correct. Would you agree?
$endgroup$
– beepboopbeepboop
Jan 24 at 1:07
$begingroup$
@KaviRamaMurthy Now, I believe the answer should be correct. Would you agree?
$endgroup$
– beepboopbeepboop
Jan 24 at 1:07
$begingroup$
You were already explained, à propos one of your earlier questions on the subject, how to deal with what you call the domain, and you were given an explicit procedure. Why not try to apply it?
$endgroup$
– Did
Jan 25 at 23:39
$begingroup$
You were already explained, à propos one of your earlier questions on the subject, how to deal with what you call the domain, and you were given an explicit procedure. Why not try to apply it?
$endgroup$
– Did
Jan 25 at 23:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We know $0 leq X leq Y$ for sure from the definition of $f_{X, Y}(x, y)$. In order to have $X + Y leq z$ we can start by thinking about the range of $Y$ values, i.e. $0 leq Y leq z$. Then given $Y$, $X$ mustlie in the range $0 leq X leq z - Y$. Combining this with the very first condition we have $0 leq X leq min(Y, z - Y)$. Therefore
$$F_Z(z) = mathbb{P}(Z leq z) = mathbb{P}(X + Y leq z) = int_{y = 0}^{y = z}int_{x = 0}^{x = min(y, z - y)}f_{X, Y}(x, y) thinspace dx thinspace dy.$$
Integrating this gives
$$F_Z(z) = int_{y = 0}^{y = z}int_{x = 0}^{x = min(y, z - y)}e^{-y} thinspace dx thinspace dy = int_{y = 0}^{y = z} e^{-y}min(y, z - y) dy.$$
When $min(y, z - y) = y Leftrightarrow y < z - y Leftrightarrow y < z/2$, so we break the integral into regions:
$$F_Z(z) = int_{0}^{z/2} ye^{-y} dy + int_{z/2}^{z} (z - y)e^{-y} dy = int_{0}^{z/2} ye^{-y} dy + zint_{z/2}^{z} e^{-y} dy - int_{z/2}^{z} ye^{-y} dy$$
And solving:
$$F_Z(z) = left[-(y + 1)e^{-y} right]_{y = 0}^{y = z/2} + z[-e^{-y}]_{y = z/2}^{y = z} - left[-(y + 1)e^{-y} right]_{y = z/2}^{y = z}$$
$$F_Z(z) = -frac{z}{2}e^{-z/2} - e^{-z/2} + e^{-0} - ze^{-z} + ze^{-z/2} + ze^{-z} + e^{-z} - frac{z}{2}e^{-z/2} - e^{-z/2}$$
$$F_Z(z) = 1 + e^{-z} - 2e^{-z/2} thinspace (text{for } 0 leq z, 0 text{ otherwise})$$
(we can check that when $F_Z(0) = 0$ and as $z rightarrow infty$, $F_Z(z) rightarrow 1$.
Differentiating gives
$$f_Z(z) = -e^{-z} + e^{-z/2} thinspace (text{for } 0 leq z, 0 text{ otherwise})$$
answered Jan 23 at 22:32
AlexAlex
709412
$endgroup$
add a comment |
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1 Answer
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$begingroup$
We know $0 leq X leq Y$ for sure from the definition of $f_{X, Y}(x, y)$. In order to have $X + Y leq z$ we can start by thinking about the range of $Y$ values, i.e. $0 leq Y leq z$. Then given $Y$, $X$ mustlie in the range $0 leq X leq z - Y$. Combining this with the very first condition we have $0 leq X leq min(Y, z - Y)$. Therefore
$$F_Z(z) = mathbb{P}(Z leq z) = mathbb{P}(X + Y leq z) = int_{y = 0}^{y = z}int_{x = 0}^{x = min(y, z - y)}f_{X, Y}(x, y) thinspace dx thinspace dy.$$
Integrating this gives
$$F_Z(z) = int_{y = 0}^{y = z}int_{x = 0}^{x = min(y, z - y)}e^{-y} thinspace dx thinspace dy = int_{y = 0}^{y = z} e^{-y}min(y, z - y) dy.$$
When $min(y, z - y) = y Leftrightarrow y < z - y Leftrightarrow y < z/2$, so we break the integral into regions:
$$F_Z(z) = int_{0}^{z/2} ye^{-y} dy + int_{z/2}^{z} (z - y)e^{-y} dy = int_{0}^{z/2} ye^{-y} dy + zint_{z/2}^{z} e^{-y} dy - int_{z/2}^{z} ye^{-y} dy$$
And solving:
$$F_Z(z) = left[-(y + 1)e^{-y} right]_{y = 0}^{y = z/2} + z[-e^{-y}]_{y = z/2}^{y = z} - left[-(y + 1)e^{-y} right]_{y = z/2}^{y = z}$$
$$F_Z(z) = -frac{z}{2}e^{-z/2} - e^{-z/2} + e^{-0} - ze^{-z} + ze^{-z/2} + ze^{-z} + e^{-z} - frac{z}{2}e^{-z/2} - e^{-z/2}$$
$$F_Z(z) = 1 + e^{-z} - 2e^{-z/2} thinspace (text{for } 0 leq z, 0 text{ otherwise})$$
(we can check that when $F_Z(0) = 0$ and as $z rightarrow infty$, $F_Z(z) rightarrow 1$.
Differentiating gives
$$f_Z(z) = -e^{-z} + e^{-z/2} thinspace (text{for } 0 leq z, 0 text{ otherwise})$$
answered Jan 23 at 22:32
AlexAlex
709412
$endgroup$
add a comment |
$begingroup$
We know $0 leq X leq Y$ for sure from the definition of $f_{X, Y}(x, y)$. In order to have $X + Y leq z$ we can start by thinking about the range of $Y$ values, i.e. $0 leq Y leq z$. Then given $Y$, $X$ mustlie in the range $0 leq X leq z - Y$. Combining this with the very first condition we have $0 leq X leq min(Y, z - Y)$. Therefore
$$F_Z(z) = mathbb{P}(Z leq z) = mathbb{P}(X + Y leq z) = int_{y = 0}^{y = z}int_{x = 0}^{x = min(y, z - y)}f_{X, Y}(x, y) thinspace dx thinspace dy.$$
Integrating this gives
$$F_Z(z) = int_{y = 0}^{y = z}int_{x = 0}^{x = min(y, z - y)}e^{-y} thinspace dx thinspace dy = int_{y = 0}^{y = z} e^{-y}min(y, z - y) dy.$$
When $min(y, z - y) = y Leftrightarrow y < z - y Leftrightarrow y < z/2$, so we break the integral into regions:
$$F_Z(z) = int_{0}^{z/2} ye^{-y} dy + int_{z/2}^{z} (z - y)e^{-y} dy = int_{0}^{z/2} ye^{-y} dy + zint_{z/2}^{z} e^{-y} dy - int_{z/2}^{z} ye^{-y} dy$$
And solving:
$$F_Z(z) = left[-(y + 1)e^{-y} right]_{y = 0}^{y = z/2} + z[-e^{-y}]_{y = z/2}^{y = z} - left[-(y + 1)e^{-y} right]_{y = z/2}^{y = z}$$
$$F_Z(z) = -frac{z}{2}e^{-z/2} - e^{-z/2} + e^{-0} - ze^{-z} + ze^{-z/2} + ze^{-z} + e^{-z} - frac{z}{2}e^{-z/2} - e^{-z/2}$$
$$F_Z(z) = 1 + e^{-z} - 2e^{-z/2} thinspace (text{for } 0 leq z, 0 text{ otherwise})$$
(we can check that when $F_Z(0) = 0$ and as $z rightarrow infty$, $F_Z(z) rightarrow 1$.
Differentiating gives
$$f_Z(z) = -e^{-z} + e^{-z/2} thinspace (text{for } 0 leq z, 0 text{ otherwise})$$
answered Jan 23 at 22:32
AlexAlex
709412
$endgroup$
add a comment |
$begingroup$
We know $0 leq X leq Y$ for sure from the definition of $f_{X, Y}(x, y)$. In order to have $X + Y leq z$ we can start by thinking about the range of $Y$ values, i.e. $0 leq Y leq z$. Then given $Y$, $X$ mustlie in the range $0 leq X leq z - Y$. Combining this with the very first condition we have $0 leq X leq min(Y, z - Y)$. Therefore
$$F_Z(z) = mathbb{P}(Z leq z) = mathbb{P}(X + Y leq z) = int_{y = 0}^{y = z}int_{x = 0}^{x = min(y, z - y)}f_{X, Y}(x, y) thinspace dx thinspace dy.$$
Integrating this gives
$$F_Z(z) = int_{y = 0}^{y = z}int_{x = 0}^{x = min(y, z - y)}e^{-y} thinspace dx thinspace dy = int_{y = 0}^{y = z} e^{-y}min(y, z - y) dy.$$
When $min(y, z - y) = y Leftrightarrow y < z - y Leftrightarrow y < z/2$, so we break the integral into regions:
$$F_Z(z) = int_{0}^{z/2} ye^{-y} dy + int_{z/2}^{z} (z - y)e^{-y} dy = int_{0}^{z/2} ye^{-y} dy + zint_{z/2}^{z} e^{-y} dy - int_{z/2}^{z} ye^{-y} dy$$
And solving:
$$F_Z(z) = left[-(y + 1)e^{-y} right]_{y = 0}^{y = z/2} + z[-e^{-y}]_{y = z/2}^{y = z} - left[-(y + 1)e^{-y} right]_{y = z/2}^{y = z}$$
$$F_Z(z) = -frac{z}{2}e^{-z/2} - e^{-z/2} + e^{-0} - ze^{-z} + ze^{-z/2} + ze^{-z} + e^{-z} - frac{z}{2}e^{-z/2} - e^{-z/2}$$
$$F_Z(z) = 1 + e^{-z} - 2e^{-z/2} thinspace (text{for } 0 leq z, 0 text{ otherwise})$$
(we can check that when $F_Z(0) = 0$ and as $z rightarrow infty$, $F_Z(z) rightarrow 1$.
Differentiating gives
$$f_Z(z) = -e^{-z} + e^{-z/2} thinspace (text{for } 0 leq z, 0 text{ otherwise})$$
answered Jan 23 at 22:32
AlexAlex
709412
$endgroup$
We know $0 leq X leq Y$ for sure from the definition of $f_{X, Y}(x, y)$. In order to have $X + Y leq z$ we can start by thinking about the range of $Y$ values, i.e. $0 leq Y leq z$. Then given $Y$, $X$ mustlie in the range $0 leq X leq z - Y$. Combining this with the very first condition we have $0 leq X leq min(Y, z - Y)$. Therefore
$$F_Z(z) = mathbb{P}(Z leq z) = mathbb{P}(X + Y leq z) = int_{y = 0}^{y = z}int_{x = 0}^{x = min(y, z - y)}f_{X, Y}(x, y) thinspace dx thinspace dy.$$
Integrating this gives
$$F_Z(z) = int_{y = 0}^{y = z}int_{x = 0}^{x = min(y, z - y)}e^{-y} thinspace dx thinspace dy = int_{y = 0}^{y = z} e^{-y}min(y, z - y) dy.$$
When $min(y, z - y) = y Leftrightarrow y < z - y Leftrightarrow y < z/2$, so we break the integral into regions:
$$F_Z(z) = int_{0}^{z/2} ye^{-y} dy + int_{z/2}^{z} (z - y)e^{-y} dy = int_{0}^{z/2} ye^{-y} dy + zint_{z/2}^{z} e^{-y} dy - int_{z/2}^{z} ye^{-y} dy$$
And solving:
$$F_Z(z) = left[-(y + 1)e^{-y} right]_{y = 0}^{y = z/2} + z[-e^{-y}]_{y = z/2}^{y = z} - left[-(y + 1)e^{-y} right]_{y = z/2}^{y = z}$$
$$F_Z(z) = -frac{z}{2}e^{-z/2} - e^{-z/2} + e^{-0} - ze^{-z} + ze^{-z/2} + ze^{-z} + e^{-z} - frac{z}{2}e^{-z/2} - e^{-z/2}$$
$$F_Z(z) = 1 + e^{-z} - 2e^{-z/2} thinspace (text{for } 0 leq z, 0 text{ otherwise})$$
(we can check that when $F_Z(0) = 0$ and as $z rightarrow infty$, $F_Z(z) rightarrow 1$.
Differentiating gives
$$f_Z(z) = -e^{-z} + e^{-z/2} thinspace (text{for } 0 leq z, 0 text{ otherwise})$$
answered Jan 23 at 22:32
AlexAlex
709412
answered Jan 23 at 22:32
AlexAlex
709412
answered Jan 23 at 22:32
AlexAlex
709412
answered Jan 23 at 22:32
AlexAlex
709412
709412
add a comment |
add a comment |
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2
$begingroup$
$f_Z(z)$ msut be a function of $z$ alone. It cannot depend on $y$.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 23:34
$begingroup$
@KaviRamaMurthy Thanks, I fixed that.
$endgroup$
– beepboopbeepboop
Jan 24 at 1:05
$begingroup$
@KaviRamaMurthy Now, I believe the answer should be correct. Would you agree?
$endgroup$
– beepboopbeepboop
Jan 24 at 1:07
$begingroup$
You were already explained, à propos one of your earlier questions on the subject, how to deal with what you call the domain, and you were given an explicit procedure. Why not try to apply it?
$endgroup$
– Did
Jan 25 at 23:39