Let $Xsim exp(lambda=1)$ and $Y sim U(1,2) $ two independent random variables; what is the joint distribution...
$begingroup$
Let $Xsim exp(lambda=1)$ and $Y sim U(1,2) $ two independent random variables; what is the joint distribution function? Compute $P(X>Y)$
I started by calculating:
$$ f(x) = begin{cases}
e^{-x} & xgeq 0\
0 &x < 0
end{cases} $$
$$ f(y) = begin{cases}
1 & 1leq yleq 2\
0 & text{otherwise}
end{cases} $$
Now I am not sure about the joint density. I do know that the fact that X,Y are independent means that I should some how multiply their densities, though I am not so sure what would be the product range.
In addition, I am not sure how I can from that derive $P(X>Y)$. Again, I have some clue about a 2-d integral but I can't figure out what's the logic behind it, or how to even define it.
Thanks!
probability probability-distributions random-variables
edited Jan 14 at 17:17
Song
18.6k21651
asked Jan 14 at 16:47
superuser123superuser123
48628
$endgroup$
add a comment |
$begingroup$
Let $Xsim exp(lambda=1)$ and $Y sim U(1,2) $ two independent random variables; what is the joint distribution function? Compute $P(X>Y)$
I started by calculating:
$$ f(x) = begin{cases}
e^{-x} & xgeq 0\
0 &x < 0
end{cases} $$
$$ f(y) = begin{cases}
1 & 1leq yleq 2\
0 & text{otherwise}
end{cases} $$
Now I am not sure about the joint density. I do know that the fact that X,Y are independent means that I should some how multiply their densities, though I am not so sure what would be the product range.
In addition, I am not sure how I can from that derive $P(X>Y)$. Again, I have some clue about a 2-d integral but I can't figure out what's the logic behind it, or how to even define it.
Thanks!
probability probability-distributions random-variables
edited Jan 14 at 17:17
Song
18.6k21651
asked Jan 14 at 16:47
superuser123superuser123
48628
$endgroup$
add a comment |
$begingroup$
Let $Xsim exp(lambda=1)$ and $Y sim U(1,2) $ two independent random variables; what is the joint distribution function? Compute $P(X>Y)$
I started by calculating:
$$ f(x) = begin{cases}
e^{-x} & xgeq 0\
0 &x < 0
end{cases} $$
$$ f(y) = begin{cases}
1 & 1leq yleq 2\
0 & text{otherwise}
end{cases} $$
Now I am not sure about the joint density. I do know that the fact that X,Y are independent means that I should some how multiply their densities, though I am not so sure what would be the product range.
In addition, I am not sure how I can from that derive $P(X>Y)$. Again, I have some clue about a 2-d integral but I can't figure out what's the logic behind it, or how to even define it.
Thanks!
probability probability-distributions random-variables
edited Jan 14 at 17:17
Song
18.6k21651
asked Jan 14 at 16:47
superuser123superuser123
48628
$endgroup$
Let $Xsim exp(lambda=1)$ and $Y sim U(1,2) $ two independent random variables; what is the joint distribution function? Compute $P(X>Y)$
I started by calculating:
$$ f(x) = begin{cases}
e^{-x} & xgeq 0\
0 &x < 0
end{cases} $$
$$ f(y) = begin{cases}
1 & 1leq yleq 2\
0 & text{otherwise}
end{cases} $$
Now I am not sure about the joint density. I do know that the fact that X,Y are independent means that I should some how multiply their densities, though I am not so sure what would be the product range.
In addition, I am not sure how I can from that derive $P(X>Y)$. Again, I have some clue about a 2-d integral but I can't figure out what's the logic behind it, or how to even define it.
Thanks!
probability probability-distributions random-variables
probability probability-distributions random-variables
edited Jan 14 at 17:17
Song
18.6k21651
asked Jan 14 at 16:47
superuser123superuser123
48628
edited Jan 14 at 17:17
Song
18.6k21651
asked Jan 14 at 16:47
superuser123superuser123
48628
edited Jan 14 at 17:17
Song
18.6k21651
edited Jan 14 at 17:17
Song
18.6k21651
edited Jan 14 at 17:17
Song
18.6k21651
18.6k21651
asked Jan 14 at 16:47
superuser123superuser123
48628
asked Jan 14 at 16:47
superuser123superuser123
48628
asked Jan 14 at 16:47
superuser123superuser123
48628
48628
add a comment |
add a comment |
2 Answers
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$begingroup$
Consider that
$$P(X>Y)=Eleft[mathbf1_{X>Y}right]$$
, where $mathbf1_{X>Y}$ equals $1$ if $X>Y$, and equals $0$ otherwise.
And for any (measurable) function $g$ of $(X,Y)$, courtesy of this theorem, we have
$$Eleft[g(X,Y)right]=iint g(x,y)f_{X,Y}(x,y),mathrm{d}x,mathrm{d}y$$
, where $f_{X,Y}$ is the joint density of $(X,Y)$.
You are right that independence of $X$ and $Y$ implies that $f_{X,Y}$ is just the product of the marginal densities of $X$ and $Y$. So you have to evaluate
begin{align}
P(X>Y)&=iint mathbf1_{x>y},e^{-x}mathbf1_{x>0}mathbf1_{1<y<2},mathrm{d}x,mathrm{d}y
\&=iint mathbf1_{x>y,,x>0,1<y<2},e^{-x},mathrm{d}x,mathrm{d}y
end{align}
answered Jan 14 at 17:16
StubbornAtomStubbornAtom
6,30831340
$endgroup$
add a comment |
$begingroup$
For the first part the joint distribution is $e^{-x}$ for the range $xgeq 0$ and $1leq yleq 2$, and 0 otherwise.
For the second part
$$text{Pr}[X>Y]=int_{y=1}^2text{Pr}[X>y],dy=int_{1}^2e^{-y},dy$$
answered Jan 14 at 17:40
BlackMathBlackMath
31518
$endgroup$
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
$begingroup$
Consider that
$$P(X>Y)=Eleft[mathbf1_{X>Y}right]$$
, where $mathbf1_{X>Y}$ equals $1$ if $X>Y$, and equals $0$ otherwise.
And for any (measurable) function $g$ of $(X,Y)$, courtesy of this theorem, we have
$$Eleft[g(X,Y)right]=iint g(x,y)f_{X,Y}(x,y),mathrm{d}x,mathrm{d}y$$
, where $f_{X,Y}$ is the joint density of $(X,Y)$.
You are right that independence of $X$ and $Y$ implies that $f_{X,Y}$ is just the product of the marginal densities of $X$ and $Y$. So you have to evaluate
begin{align}
P(X>Y)&=iint mathbf1_{x>y},e^{-x}mathbf1_{x>0}mathbf1_{1<y<2},mathrm{d}x,mathrm{d}y
\&=iint mathbf1_{x>y,,x>0,1<y<2},e^{-x},mathrm{d}x,mathrm{d}y
end{align}
answered Jan 14 at 17:16
StubbornAtomStubbornAtom
6,30831340
$endgroup$
add a comment |
$begingroup$
Consider that
$$P(X>Y)=Eleft[mathbf1_{X>Y}right]$$
, where $mathbf1_{X>Y}$ equals $1$ if $X>Y$, and equals $0$ otherwise.
And for any (measurable) function $g$ of $(X,Y)$, courtesy of this theorem, we have
$$Eleft[g(X,Y)right]=iint g(x,y)f_{X,Y}(x,y),mathrm{d}x,mathrm{d}y$$
, where $f_{X,Y}$ is the joint density of $(X,Y)$.
You are right that independence of $X$ and $Y$ implies that $f_{X,Y}$ is just the product of the marginal densities of $X$ and $Y$. So you have to evaluate
begin{align}
P(X>Y)&=iint mathbf1_{x>y},e^{-x}mathbf1_{x>0}mathbf1_{1<y<2},mathrm{d}x,mathrm{d}y
\&=iint mathbf1_{x>y,,x>0,1<y<2},e^{-x},mathrm{d}x,mathrm{d}y
end{align}
answered Jan 14 at 17:16
StubbornAtomStubbornAtom
6,30831340
$endgroup$
add a comment |
$begingroup$
Consider that
$$P(X>Y)=Eleft[mathbf1_{X>Y}right]$$
, where $mathbf1_{X>Y}$ equals $1$ if $X>Y$, and equals $0$ otherwise.
And for any (measurable) function $g$ of $(X,Y)$, courtesy of this theorem, we have
$$Eleft[g(X,Y)right]=iint g(x,y)f_{X,Y}(x,y),mathrm{d}x,mathrm{d}y$$
, where $f_{X,Y}$ is the joint density of $(X,Y)$.
You are right that independence of $X$ and $Y$ implies that $f_{X,Y}$ is just the product of the marginal densities of $X$ and $Y$. So you have to evaluate
begin{align}
P(X>Y)&=iint mathbf1_{x>y},e^{-x}mathbf1_{x>0}mathbf1_{1<y<2},mathrm{d}x,mathrm{d}y
\&=iint mathbf1_{x>y,,x>0,1<y<2},e^{-x},mathrm{d}x,mathrm{d}y
end{align}
answered Jan 14 at 17:16
StubbornAtomStubbornAtom
6,30831340
$endgroup$
Consider that
$$P(X>Y)=Eleft[mathbf1_{X>Y}right]$$
, where $mathbf1_{X>Y}$ equals $1$ if $X>Y$, and equals $0$ otherwise.
And for any (measurable) function $g$ of $(X,Y)$, courtesy of this theorem, we have
$$Eleft[g(X,Y)right]=iint g(x,y)f_{X,Y}(x,y),mathrm{d}x,mathrm{d}y$$
, where $f_{X,Y}$ is the joint density of $(X,Y)$.
You are right that independence of $X$ and $Y$ implies that $f_{X,Y}$ is just the product of the marginal densities of $X$ and $Y$. So you have to evaluate
begin{align}
P(X>Y)&=iint mathbf1_{x>y},e^{-x}mathbf1_{x>0}mathbf1_{1<y<2},mathrm{d}x,mathrm{d}y
\&=iint mathbf1_{x>y,,x>0,1<y<2},e^{-x},mathrm{d}x,mathrm{d}y
end{align}
answered Jan 14 at 17:16
StubbornAtomStubbornAtom
6,30831340
edited Jan 14 at 17:26
answered Jan 14 at 17:16
StubbornAtomStubbornAtom
6,30831340
answered Jan 14 at 17:16
StubbornAtomStubbornAtom
6,30831340
answered Jan 14 at 17:16
StubbornAtomStubbornAtom
6,30831340
6,30831340
add a comment |
add a comment |
$begingroup$
For the first part the joint distribution is $e^{-x}$ for the range $xgeq 0$ and $1leq yleq 2$, and 0 otherwise.
For the second part
$$text{Pr}[X>Y]=int_{y=1}^2text{Pr}[X>y],dy=int_{1}^2e^{-y},dy$$
answered Jan 14 at 17:40
BlackMathBlackMath
31518
$endgroup$
add a comment |
$begingroup$
For the first part the joint distribution is $e^{-x}$ for the range $xgeq 0$ and $1leq yleq 2$, and 0 otherwise.
For the second part
$$text{Pr}[X>Y]=int_{y=1}^2text{Pr}[X>y],dy=int_{1}^2e^{-y},dy$$
answered Jan 14 at 17:40
BlackMathBlackMath
31518
$endgroup$
add a comment |
$begingroup$
For the first part the joint distribution is $e^{-x}$ for the range $xgeq 0$ and $1leq yleq 2$, and 0 otherwise.
For the second part
$$text{Pr}[X>Y]=int_{y=1}^2text{Pr}[X>y],dy=int_{1}^2e^{-y},dy$$
answered Jan 14 at 17:40
BlackMathBlackMath
31518
$endgroup$
For the first part the joint distribution is $e^{-x}$ for the range $xgeq 0$ and $1leq yleq 2$, and 0 otherwise.
For the second part
$$text{Pr}[X>Y]=int_{y=1}^2text{Pr}[X>y],dy=int_{1}^2e^{-y},dy$$
answered Jan 14 at 17:40
BlackMathBlackMath
31518
answered Jan 14 at 17:40
BlackMathBlackMath
31518
answered Jan 14 at 17:40
BlackMathBlackMath
31518
answered Jan 14 at 17:40
BlackMathBlackMath
31518
31518
add a comment |
add a comment |
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