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Random Variable X and Y has a joint probability density function. Find $P(x | y)(x | y)$
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Random Variable X and Y has a joint probability density function.
$$f_X, Y (x, y) =begincases
c(x + 3y)& 3 leq x leq 7, 4 leq y leq10\
0 & textotherwise \
endcases
$$
(a) Find $f_X (x | y)$
(b) $P(x leq 5 | Y = 9)$
My attempt:
$f_X (x | y) = fracf_X, Y(x, y)f_Y(y)$
$$f_Y(y) = cint_3^7(x+3y)dx = cleft(fracx^22 + 3xy right)bigg|_3^7 = c(20+12y)$$
for $f_Y(y)$ has support $4 leq y leq 10$, 0 otherwise
$$P_X (x | y) = fracc(x+3y)c(20+12y) = frac(x+3y)(20+12y)$$
$P_X (x | y)$ has support ...
(b)
$f_X (x | y = 9) = fracx + 3 cdot 920 + 12 cdot 9 = fracx+27128$
$$P(x leq 5 | Y = 9) = int_3^5 f_X (x | y = 9)dx = int_3^5 fracx+27128 = int_3^5 left(x/128 + 27/128 right)dx = left(fracx^2256 + frac27128x right)bigg|_3^5 = frac3164$$
Not sure if I'm right, and a question if it were the joint pdf as
$$f_X, Y (x, y) =begincases
c(x + 3y)& 3 leq x leq y, 4 leq y leq10\
0 & textotherwise \
endcases
$$
How would the integral for the first question look like?
probability probability-distributions
asked Jul 19 at 3:14
Bas bas
39611
add a comment |Â
up vote
0
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Random Variable X and Y has a joint probability density function.
$$f_X, Y (x, y) =begincases
c(x + 3y)& 3 leq x leq 7, 4 leq y leq10\
0 & textotherwise \
endcases
$$
(a) Find $f_X (x | y)$
(b) $P(x leq 5 | Y = 9)$
My attempt:
$f_X (x | y) = fracf_X, Y(x, y)f_Y(y)$
$$f_Y(y) = cint_3^7(x+3y)dx = cleft(fracx^22 + 3xy right)bigg|_3^7 = c(20+12y)$$
for $f_Y(y)$ has support $4 leq y leq 10$, 0 otherwise
$$P_X (x | y) = fracc(x+3y)c(20+12y) = frac(x+3y)(20+12y)$$
$P_X (x | y)$ has support ...
(b)
$f_X (x | y = 9) = fracx + 3 cdot 920 + 12 cdot 9 = fracx+27128$
$$P(x leq 5 | Y = 9) = int_3^5 f_X (x | y = 9)dx = int_3^5 fracx+27128 = int_3^5 left(x/128 + 27/128 right)dx = left(fracx^2256 + frac27128x right)bigg|_3^5 = frac3164$$
Not sure if I'm right, and a question if it were the joint pdf as
$$f_X, Y (x, y) =begincases
c(x + 3y)& 3 leq x leq y, 4 leq y leq10\
0 & textotherwise \
endcases
$$
How would the integral for the first question look like?
probability probability-distributions
asked Jul 19 at 3:14
Bas bas
39611
The first part looks correct to me.
– chandresh
Jul 19 at 3:47
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Random Variable X and Y has a joint probability density function.
$$f_X, Y (x, y) =begincases
c(x + 3y)& 3 leq x leq 7, 4 leq y leq10\
0 & textotherwise \
endcases
$$
(a) Find $f_X (x | y)$
(b) $P(x leq 5 | Y = 9)$
My attempt:
$f_X (x | y) = fracf_X, Y(x, y)f_Y(y)$
$$f_Y(y) = cint_3^7(x+3y)dx = cleft(fracx^22 + 3xy right)bigg|_3^7 = c(20+12y)$$
for $f_Y(y)$ has support $4 leq y leq 10$, 0 otherwise
$$P_X (x | y) = fracc(x+3y)c(20+12y) = frac(x+3y)(20+12y)$$
$P_X (x | y)$ has support ...
(b)
$f_X (x | y = 9) = fracx + 3 cdot 920 + 12 cdot 9 = fracx+27128$
$$P(x leq 5 | Y = 9) = int_3^5 f_X (x | y = 9)dx = int_3^5 fracx+27128 = int_3^5 left(x/128 + 27/128 right)dx = left(fracx^2256 + frac27128x right)bigg|_3^5 = frac3164$$
Not sure if I'm right, and a question if it were the joint pdf as
$$f_X, Y (x, y) =begincases
c(x + 3y)& 3 leq x leq y, 4 leq y leq10\
0 & textotherwise \
endcases
$$
How would the integral for the first question look like?
probability probability-distributions
asked Jul 19 at 3:14
Bas bas
39611
Random Variable X and Y has a joint probability density function.
$$f_X, Y (x, y) =begincases
c(x + 3y)& 3 leq x leq 7, 4 leq y leq10\
0 & textotherwise \
endcases
$$
(a) Find $f_X (x | y)$
(b) $P(x leq 5 | Y = 9)$
My attempt:
$f_X (x | y) = fracf_X, Y(x, y)f_Y(y)$
$$f_Y(y) = cint_3^7(x+3y)dx = cleft(fracx^22 + 3xy right)bigg|_3^7 = c(20+12y)$$
for $f_Y(y)$ has support $4 leq y leq 10$, 0 otherwise
$$P_X (x | y) = fracc(x+3y)c(20+12y) = frac(x+3y)(20+12y)$$
$P_X (x | y)$ has support ...
(b)
$f_X (x | y = 9) = fracx + 3 cdot 920 + 12 cdot 9 = fracx+27128$
$$P(x leq 5 | Y = 9) = int_3^5 f_X (x | y = 9)dx = int_3^5 fracx+27128 = int_3^5 left(x/128 + 27/128 right)dx = left(fracx^2256 + frac27128x right)bigg|_3^5 = frac3164$$
Not sure if I'm right, and a question if it were the joint pdf as
$$f_X, Y (x, y) =begincases
c(x + 3y)& 3 leq x leq y, 4 leq y leq10\
0 & textotherwise \
endcases
$$
How would the integral for the first question look like?
probability probability-distributions
asked Jul 19 at 3:14
Bas bas
39611
edited Jul 19 at 3:52
asked Jul 19 at 3:14
Bas bas
39611
asked Jul 19 at 3:14
Bas bas
39611
asked Jul 19 at 3:14
Bas bas
39611
39611
The first part looks correct to me.
– chandresh
Jul 19 at 3:47
add a comment |Â
The first part looks correct to me.
– chandresh
Jul 19 at 3:47
The first part looks correct to me.
– chandresh
Jul 19 at 3:47
The first part looks correct to me.
– chandresh
Jul 19 at 3:47
add a comment |Â
1 Answer
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0
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You are correct. Â Don't forget, however, to indicate the supports.
 (Well, assuming $P_Xmid Y$ is conditional probability density function; why did they use $P$ rather than $f$?)
If instead $f_X,Y(x,y) = c(x+2y),mathbf 1_3<x<y, 4<y<10$, then $$f_Y(y)~=int_3^y c(x+2y)mathbf 1_4<y<10,mathsf d x \= c(tfrac 52y^2-6y-tfrac 92),mathbf 1_4<y<10$$
answered Jul 19 at 3:49
Graham Kemp
80.1k43275
What would I put for the support for $f_Y(x|y)$?
– Bas bas
Jul 19 at 3:54
The same as the support for the joint density function. @Basbas .
– Graham Kemp
Jul 19 at 3:56
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You are correct. Â Don't forget, however, to indicate the supports.
 (Well, assuming $P_Xmid Y$ is conditional probability density function; why did they use $P$ rather than $f$?)
If instead $f_X,Y(x,y) = c(x+2y),mathbf 1_3<x<y, 4<y<10$, then $$f_Y(y)~=int_3^y c(x+2y)mathbf 1_4<y<10,mathsf d x \= c(tfrac 52y^2-6y-tfrac 92),mathbf 1_4<y<10$$
answered Jul 19 at 3:49
Graham Kemp
80.1k43275
What would I put for the support for $f_Y(x|y)$?
– Bas bas
Jul 19 at 3:54
The same as the support for the joint density function. @Basbas .
– Graham Kemp
Jul 19 at 3:56
add a comment |Â
up vote
0
down vote
accepted
You are correct. Â Don't forget, however, to indicate the supports.
 (Well, assuming $P_Xmid Y$ is conditional probability density function; why did they use $P$ rather than $f$?)
If instead $f_X,Y(x,y) = c(x+2y),mathbf 1_3<x<y, 4<y<10$, then $$f_Y(y)~=int_3^y c(x+2y)mathbf 1_4<y<10,mathsf d x \= c(tfrac 52y^2-6y-tfrac 92),mathbf 1_4<y<10$$
answered Jul 19 at 3:49
Graham Kemp
80.1k43275
What would I put for the support for $f_Y(x|y)$?
– Bas bas
Jul 19 at 3:54
The same as the support for the joint density function. @Basbas .
– Graham Kemp
Jul 19 at 3:56
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You are correct. Â Don't forget, however, to indicate the supports.
 (Well, assuming $P_Xmid Y$ is conditional probability density function; why did they use $P$ rather than $f$?)
If instead $f_X,Y(x,y) = c(x+2y),mathbf 1_3<x<y, 4<y<10$, then $$f_Y(y)~=int_3^y c(x+2y)mathbf 1_4<y<10,mathsf d x \= c(tfrac 52y^2-6y-tfrac 92),mathbf 1_4<y<10$$
answered Jul 19 at 3:49
Graham Kemp
80.1k43275
You are correct. Â Don't forget, however, to indicate the supports.
 (Well, assuming $P_Xmid Y$ is conditional probability density function; why did they use $P$ rather than $f$?)
If instead $f_X,Y(x,y) = c(x+2y),mathbf 1_3<x<y, 4<y<10$, then $$f_Y(y)~=int_3^y c(x+2y)mathbf 1_4<y<10,mathsf d x \= c(tfrac 52y^2-6y-tfrac 92),mathbf 1_4<y<10$$
answered Jul 19 at 3:49
Graham Kemp
80.1k43275
answered Jul 19 at 3:49
Graham Kemp
80.1k43275
answered Jul 19 at 3:49
Graham Kemp
80.1k43275
answered Jul 19 at 3:49
Graham Kemp
80.1k43275
80.1k43275
What would I put for the support for $f_Y(x|y)$?
– Bas bas
Jul 19 at 3:54
The same as the support for the joint density function. @Basbas .
– Graham Kemp
Jul 19 at 3:56
add a comment |Â
What would I put for the support for $f_Y(x|y)$?
– Bas bas
Jul 19 at 3:54
The same as the support for the joint density function. @Basbas .
– Graham Kemp
Jul 19 at 3:56
What would I put for the support for $f_Y(x|y)$?
– Bas bas
Jul 19 at 3:54
What would I put for the support for $f_Y(x|y)$?
– Bas bas
Jul 19 at 3:54
The same as the support for the joint density function. @Basbas .
– Graham Kemp
Jul 19 at 3:56
The same as the support for the joint density function. @Basbas .
– Graham Kemp
Jul 19 at 3:56
add a comment |Â
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The first part looks correct to me.
– chandresh
Jul 19 at 3:47