finding probability of joint distribution.
I am given a joint density function of two random variables $X$ and $Y$ which is as follows:
$$
f(x,y) = left{
begin{array}{ll}
frac{1}{210}(2x+y) & quad 2<x <6, 0<y<5 \
0 & quadtext{otherwise}
end{array}
right.
$$
Now I want to calculate the probability $P(X+Y < 4)$. Will I do it like this?:
$$ int_{0}^{4}int_{2}^{4} frac{1}{210}(2x+y) dxdy $$
probability probability-distributions
edited Dec 26 '18 at 18:17
APC89
1,926418
asked Dec 26 '18 at 18:16
Ahmad Qayyum
555
add a comment |
I am given a joint density function of two random variables $X$ and $Y$ which is as follows:
$$
f(x,y) = left{
begin{array}{ll}
frac{1}{210}(2x+y) & quad 2<x <6, 0<y<5 \
0 & quadtext{otherwise}
end{array}
right.
$$
Now I want to calculate the probability $P(X+Y < 4)$. Will I do it like this?:
$$ int_{0}^{4}int_{2}^{4} frac{1}{210}(2x+y) dxdy $$
probability probability-distributions
edited Dec 26 '18 at 18:17
APC89
1,926418
asked Dec 26 '18 at 18:16
Ahmad Qayyum
555
add a comment |
I am given a joint density function of two random variables $X$ and $Y$ which is as follows:
$$
f(x,y) = left{
begin{array}{ll}
frac{1}{210}(2x+y) & quad 2<x <6, 0<y<5 \
0 & quadtext{otherwise}
end{array}
right.
$$
Now I want to calculate the probability $P(X+Y < 4)$. Will I do it like this?:
$$ int_{0}^{4}int_{2}^{4} frac{1}{210}(2x+y) dxdy $$
probability probability-distributions
edited Dec 26 '18 at 18:17
APC89
1,926418
asked Dec 26 '18 at 18:16
Ahmad Qayyum
555
I am given a joint density function of two random variables $X$ and $Y$ which is as follows:
$$
f(x,y) = left{
begin{array}{ll}
frac{1}{210}(2x+y) & quad 2<x <6, 0<y<5 \
0 & quadtext{otherwise}
end{array}
right.
$$
Now I want to calculate the probability $P(X+Y < 4)$. Will I do it like this?:
$$ int_{0}^{4}int_{2}^{4} frac{1}{210}(2x+y) dxdy $$
probability probability-distributions
probability probability-distributions
edited Dec 26 '18 at 18:17
APC89
1,926418
asked Dec 26 '18 at 18:16
Ahmad Qayyum
555
edited Dec 26 '18 at 18:17
APC89
1,926418
asked Dec 26 '18 at 18:16
Ahmad Qayyum
555
edited Dec 26 '18 at 18:17
APC89
1,926418
edited Dec 26 '18 at 18:17
APC89
1,926418
edited Dec 26 '18 at 18:17
APC89
1,926418
1,926418
asked Dec 26 '18 at 18:16
Ahmad Qayyum
555
asked Dec 26 '18 at 18:16
Ahmad Qayyum
555
asked Dec 26 '18 at 18:16
Ahmad Qayyum
555
555
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
This is technically an exercise on multiple integrals, and in these cases you would realise that your job becomes easier if you can picture the desired region of integration.
Proceed step by step to focus on the domain of integration.
begin{align}
P(X+Y<4)&=Eleft[mathbf1_{X+Y<4}right]
\&=iint mathbf1_{x+y<4},f(x,y),dx,dy
\&=frac{1}{210}iint mathbf1_{x+y<4},(2x+y)mathbf1_{2<x<6,,0<y<5},dx,dy
\&=frac{1}{210}iint (2x+y),mathbf1_{x+y<4,,2<x<6,,0<y<5},dx,dy
end{align}
This is how the region ${(x,y):x+y<4,,2<x<6,,0<y<5}$ looks like:
Can you evaluate the double integral now, keeping any one of $x$ and $y$ fixed at a time depending on ease of computation ?
answered Dec 26 '18 at 18:57
StubbornAtom
5,29211138
so what will be the limits over the integrals?
– Ahmad Qayyum
Dec 26 '18 at 19:25
@AhmadQayyum You tell me.
– StubbornAtom
Dec 26 '18 at 19:26
will these be the limits?: $$ int_{0}^{4-x}int_{2}^{4} frac{1}{210}(2x+y) dxdy $$
– Ahmad Qayyum
Dec 26 '18 at 19:28
after i solved it i got the final expression as: $ frac{1}{210}(x-16)(x-4) $. Is it correct or do I have to solve it further?
– Ahmad Qayyum
Dec 26 '18 at 19:40
1
@AhmadQayyum Sorry, it is rather $$int_2^4 int_0^{4-x}...,dy,dx$$ Since you are keeping $x$ free and $y$ dependent on $x$, you have to integrate wrt $y$ first and then $x$.
– StubbornAtom
Dec 26 '18 at 19:49
|
show 1 more comment
No, that integral gives $mathbb{P}(2leq Xleq 4,, 0leq Y leq 4)$. For example, this includes $(4,4)$ which clearly does not satisfy $X + Y < 4$. You're integrating over a rectangular region, you should be integrating over a triangular region. Try
$$
int_2^4int_0^{4-x} frac{1}{210}(2x+y),dy,dx
$$
answered Dec 26 '18 at 18:22
Alex
1777
You're right, edited. Thanks!
– Alex
Dec 26 '18 at 21:18
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
This is technically an exercise on multiple integrals, and in these cases you would realise that your job becomes easier if you can picture the desired region of integration.
Proceed step by step to focus on the domain of integration.
begin{align}
P(X+Y<4)&=Eleft[mathbf1_{X+Y<4}right]
\&=iint mathbf1_{x+y<4},f(x,y),dx,dy
\&=frac{1}{210}iint mathbf1_{x+y<4},(2x+y)mathbf1_{2<x<6,,0<y<5},dx,dy
\&=frac{1}{210}iint (2x+y),mathbf1_{x+y<4,,2<x<6,,0<y<5},dx,dy
end{align}
This is how the region ${(x,y):x+y<4,,2<x<6,,0<y<5}$ looks like:
Can you evaluate the double integral now, keeping any one of $x$ and $y$ fixed at a time depending on ease of computation ?
answered Dec 26 '18 at 18:57
StubbornAtom
5,29211138
so what will be the limits over the integrals?
– Ahmad Qayyum
Dec 26 '18 at 19:25
@AhmadQayyum You tell me.
– StubbornAtom
Dec 26 '18 at 19:26
will these be the limits?: $$ int_{0}^{4-x}int_{2}^{4} frac{1}{210}(2x+y) dxdy $$
– Ahmad Qayyum
Dec 26 '18 at 19:28
after i solved it i got the final expression as: $ frac{1}{210}(x-16)(x-4) $. Is it correct or do I have to solve it further?
– Ahmad Qayyum
Dec 26 '18 at 19:40
1
@AhmadQayyum Sorry, it is rather $$int_2^4 int_0^{4-x}...,dy,dx$$ Since you are keeping $x$ free and $y$ dependent on $x$, you have to integrate wrt $y$ first and then $x$.
– StubbornAtom
Dec 26 '18 at 19:49
|
show 1 more comment
This is technically an exercise on multiple integrals, and in these cases you would realise that your job becomes easier if you can picture the desired region of integration.
Proceed step by step to focus on the domain of integration.
begin{align}
P(X+Y<4)&=Eleft[mathbf1_{X+Y<4}right]
\&=iint mathbf1_{x+y<4},f(x,y),dx,dy
\&=frac{1}{210}iint mathbf1_{x+y<4},(2x+y)mathbf1_{2<x<6,,0<y<5},dx,dy
\&=frac{1}{210}iint (2x+y),mathbf1_{x+y<4,,2<x<6,,0<y<5},dx,dy
end{align}
This is how the region ${(x,y):x+y<4,,2<x<6,,0<y<5}$ looks like:
Can you evaluate the double integral now, keeping any one of $x$ and $y$ fixed at a time depending on ease of computation ?
answered Dec 26 '18 at 18:57
StubbornAtom
5,29211138
so what will be the limits over the integrals?
– Ahmad Qayyum
Dec 26 '18 at 19:25
@AhmadQayyum You tell me.
– StubbornAtom
Dec 26 '18 at 19:26
will these be the limits?: $$ int_{0}^{4-x}int_{2}^{4} frac{1}{210}(2x+y) dxdy $$
– Ahmad Qayyum
Dec 26 '18 at 19:28
after i solved it i got the final expression as: $ frac{1}{210}(x-16)(x-4) $. Is it correct or do I have to solve it further?
– Ahmad Qayyum
Dec 26 '18 at 19:40
1
@AhmadQayyum Sorry, it is rather $$int_2^4 int_0^{4-x}...,dy,dx$$ Since you are keeping $x$ free and $y$ dependent on $x$, you have to integrate wrt $y$ first and then $x$.
– StubbornAtom
Dec 26 '18 at 19:49
|
show 1 more comment
This is technically an exercise on multiple integrals, and in these cases you would realise that your job becomes easier if you can picture the desired region of integration.
Proceed step by step to focus on the domain of integration.
begin{align}
P(X+Y<4)&=Eleft[mathbf1_{X+Y<4}right]
\&=iint mathbf1_{x+y<4},f(x,y),dx,dy
\&=frac{1}{210}iint mathbf1_{x+y<4},(2x+y)mathbf1_{2<x<6,,0<y<5},dx,dy
\&=frac{1}{210}iint (2x+y),mathbf1_{x+y<4,,2<x<6,,0<y<5},dx,dy
end{align}
This is how the region ${(x,y):x+y<4,,2<x<6,,0<y<5}$ looks like:
Can you evaluate the double integral now, keeping any one of $x$ and $y$ fixed at a time depending on ease of computation ?
answered Dec 26 '18 at 18:57
StubbornAtom
5,29211138
This is technically an exercise on multiple integrals, and in these cases you would realise that your job becomes easier if you can picture the desired region of integration.
Proceed step by step to focus on the domain of integration.
begin{align}
P(X+Y<4)&=Eleft[mathbf1_{X+Y<4}right]
\&=iint mathbf1_{x+y<4},f(x,y),dx,dy
\&=frac{1}{210}iint mathbf1_{x+y<4},(2x+y)mathbf1_{2<x<6,,0<y<5},dx,dy
\&=frac{1}{210}iint (2x+y),mathbf1_{x+y<4,,2<x<6,,0<y<5},dx,dy
end{align}
This is how the region ${(x,y):x+y<4,,2<x<6,,0<y<5}$ looks like:
Can you evaluate the double integral now, keeping any one of $x$ and $y$ fixed at a time depending on ease of computation ?
answered Dec 26 '18 at 18:57
StubbornAtom
5,29211138
edited Dec 26 '18 at 19:22
answered Dec 26 '18 at 18:57
StubbornAtom
5,29211138
answered Dec 26 '18 at 18:57
StubbornAtom
5,29211138
answered Dec 26 '18 at 18:57
StubbornAtom
5,29211138
5,29211138
so what will be the limits over the integrals?
– Ahmad Qayyum
Dec 26 '18 at 19:25
@AhmadQayyum You tell me.
– StubbornAtom
Dec 26 '18 at 19:26
will these be the limits?: $$ int_{0}^{4-x}int_{2}^{4} frac{1}{210}(2x+y) dxdy $$
– Ahmad Qayyum
Dec 26 '18 at 19:28
after i solved it i got the final expression as: $ frac{1}{210}(x-16)(x-4) $. Is it correct or do I have to solve it further?
– Ahmad Qayyum
Dec 26 '18 at 19:40
1
@AhmadQayyum Sorry, it is rather $$int_2^4 int_0^{4-x}...,dy,dx$$ Since you are keeping $x$ free and $y$ dependent on $x$, you have to integrate wrt $y$ first and then $x$.
– StubbornAtom
Dec 26 '18 at 19:49
|
show 1 more comment
so what will be the limits over the integrals?
– Ahmad Qayyum
Dec 26 '18 at 19:25
@AhmadQayyum You tell me.
– StubbornAtom
Dec 26 '18 at 19:26
will these be the limits?: $$ int_{0}^{4-x}int_{2}^{4} frac{1}{210}(2x+y) dxdy $$
– Ahmad Qayyum
Dec 26 '18 at 19:28
after i solved it i got the final expression as: $ frac{1}{210}(x-16)(x-4) $. Is it correct or do I have to solve it further?
– Ahmad Qayyum
Dec 26 '18 at 19:40
1
@AhmadQayyum Sorry, it is rather $$int_2^4 int_0^{4-x}...,dy,dx$$ Since you are keeping $x$ free and $y$ dependent on $x$, you have to integrate wrt $y$ first and then $x$.
– StubbornAtom
Dec 26 '18 at 19:49
so what will be the limits over the integrals?
– Ahmad Qayyum
Dec 26 '18 at 19:25
so what will be the limits over the integrals?
– Ahmad Qayyum
Dec 26 '18 at 19:25
@AhmadQayyum You tell me.
– StubbornAtom
Dec 26 '18 at 19:26
@AhmadQayyum You tell me.
– StubbornAtom
Dec 26 '18 at 19:26
will these be the limits?: $$ int_{0}^{4-x}int_{2}^{4} frac{1}{210}(2x+y) dxdy $$
– Ahmad Qayyum
Dec 26 '18 at 19:28
will these be the limits?: $$ int_{0}^{4-x}int_{2}^{4} frac{1}{210}(2x+y) dxdy $$
– Ahmad Qayyum
Dec 26 '18 at 19:28
after i solved it i got the final expression as: $ frac{1}{210}(x-16)(x-4) $. Is it correct or do I have to solve it further?
– Ahmad Qayyum
Dec 26 '18 at 19:40
after i solved it i got the final expression as: $ frac{1}{210}(x-16)(x-4) $. Is it correct or do I have to solve it further?
– Ahmad Qayyum
Dec 26 '18 at 19:40
1
1
@AhmadQayyum Sorry, it is rather $$int_2^4 int_0^{4-x}...,dy,dx$$ Since you are keeping $x$ free and $y$ dependent on $x$, you have to integrate wrt $y$ first and then $x$.
– StubbornAtom
Dec 26 '18 at 19:49
@AhmadQayyum Sorry, it is rather $$int_2^4 int_0^{4-x}...,dy,dx$$ Since you are keeping $x$ free and $y$ dependent on $x$, you have to integrate wrt $y$ first and then $x$.
– StubbornAtom
Dec 26 '18 at 19:49
|
show 1 more comment
No, that integral gives $mathbb{P}(2leq Xleq 4,, 0leq Y leq 4)$. For example, this includes $(4,4)$ which clearly does not satisfy $X + Y < 4$. You're integrating over a rectangular region, you should be integrating over a triangular region. Try
$$
int_2^4int_0^{4-x} frac{1}{210}(2x+y),dy,dx
$$
answered Dec 26 '18 at 18:22
Alex
1777
You're right, edited. Thanks!
– Alex
Dec 26 '18 at 21:18
add a comment |
No, that integral gives $mathbb{P}(2leq Xleq 4,, 0leq Y leq 4)$. For example, this includes $(4,4)$ which clearly does not satisfy $X + Y < 4$. You're integrating over a rectangular region, you should be integrating over a triangular region. Try
$$
int_2^4int_0^{4-x} frac{1}{210}(2x+y),dy,dx
$$
answered Dec 26 '18 at 18:22
Alex
1777
You're right, edited. Thanks!
– Alex
Dec 26 '18 at 21:18
add a comment |
No, that integral gives $mathbb{P}(2leq Xleq 4,, 0leq Y leq 4)$. For example, this includes $(4,4)$ which clearly does not satisfy $X + Y < 4$. You're integrating over a rectangular region, you should be integrating over a triangular region. Try
$$
int_2^4int_0^{4-x} frac{1}{210}(2x+y),dy,dx
$$
answered Dec 26 '18 at 18:22
Alex
1777
No, that integral gives $mathbb{P}(2leq Xleq 4,, 0leq Y leq 4)$. For example, this includes $(4,4)$ which clearly does not satisfy $X + Y < 4$. You're integrating over a rectangular region, you should be integrating over a triangular region. Try
$$
int_2^4int_0^{4-x} frac{1}{210}(2x+y),dy,dx
$$
answered Dec 26 '18 at 18:22
Alex
1777
edited Dec 26 '18 at 21:18
answered Dec 26 '18 at 18:22
Alex
1777
answered Dec 26 '18 at 18:22
Alex
1777
answered Dec 26 '18 at 18:22
Alex
1777
1777
You're right, edited. Thanks!
– Alex
Dec 26 '18 at 21:18
add a comment |
You're right, edited. Thanks!
– Alex
Dec 26 '18 at 21:18
You're right, edited. Thanks!
– Alex
Dec 26 '18 at 21:18
You're right, edited. Thanks!
– Alex
Dec 26 '18 at 21:18
add a comment |
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