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Find the value of $k$ for $f(x|y)$
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Let $Y$ a uniform random variable in $[0,1]$, and let $X$ a uniform variable in $[1,e^Y]$
Find $f(x|y)$ and $f_y(y)$
We know $Xsim mathcalU[1,e^Y]$ and $Ysim mathcalU[0,1]$ then by definition of uniform distribution, we have:
$f(x|y)= begincases k & 1 le x le e^y \0 & otherwise,\endcases$
and $f_y(y)=begincases t & 0 le y le 1 \0 & otherwise,\endcases$
Then, i need find the value of $t$ and $k$.
For $f_y(y)$ by definition of density function, $t=1$. But for $f(x|y)$ i'm stuck. Can someone help me?
probability
asked Jul 15 at 23:40
Orlian Prato
275
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up vote
2
down vote
favorite
Let $Y$ a uniform random variable in $[0,1]$, and let $X$ a uniform variable in $[1,e^Y]$
Find $f(x|y)$ and $f_y(y)$
We know $Xsim mathcalU[1,e^Y]$ and $Ysim mathcalU[0,1]$ then by definition of uniform distribution, we have:
$f(x|y)= begincases k & 1 le x le e^y \0 & otherwise,\endcases$
and $f_y(y)=begincases t & 0 le y le 1 \0 & otherwise,\endcases$
Then, i need find the value of $t$ and $k$.
For $f_y(y)$ by definition of density function, $t=1$. But for $f(x|y)$ i'm stuck. Can someone help me?
probability
asked Jul 15 at 23:40
Orlian Prato
275
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $Y$ a uniform random variable in $[0,1]$, and let $X$ a uniform variable in $[1,e^Y]$
Find $f(x|y)$ and $f_y(y)$
We know $Xsim mathcalU[1,e^Y]$ and $Ysim mathcalU[0,1]$ then by definition of uniform distribution, we have:
$f(x|y)= begincases k & 1 le x le e^y \0 & otherwise,\endcases$
and $f_y(y)=begincases t & 0 le y le 1 \0 & otherwise,\endcases$
Then, i need find the value of $t$ and $k$.
For $f_y(y)$ by definition of density function, $t=1$. But for $f(x|y)$ i'm stuck. Can someone help me?
probability
asked Jul 15 at 23:40
Orlian Prato
275
Let $Y$ a uniform random variable in $[0,1]$, and let $X$ a uniform variable in $[1,e^Y]$
Find $f(x|y)$ and $f_y(y)$
We know $Xsim mathcalU[1,e^Y]$ and $Ysim mathcalU[0,1]$ then by definition of uniform distribution, we have:
$f(x|y)= begincases k & 1 le x le e^y \0 & otherwise,\endcases$
and $f_y(y)=begincases t & 0 le y le 1 \0 & otherwise,\endcases$
Then, i need find the value of $t$ and $k$.
For $f_y(y)$ by definition of density function, $t=1$. But for $f(x|y)$ i'm stuck. Can someone help me?
probability
asked Jul 15 at 23:40
Orlian Prato
275
asked Jul 15 at 23:40
Orlian Prato
275
asked Jul 15 at 23:40
Orlian Prato
275
asked Jul 15 at 23:40
Orlian Prato
275
275
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
The integral of a pdf from $-infty$ to $infty$ (the total area under the curve) must be 1. Therefore, $k$ is $(e^y-1)^-1$. Similarly for $t$.
edited Jul 16 at 0:14
Graham Kemp
80.1k43275
answered Jul 15 at 23:46
NicNic8
3,7113922
@Graham Kemp oops. Thanks. :)
– NicNic8
Jul 16 at 0:24
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The integral of a pdf from $-infty$ to $infty$ (the total area under the curve) must be 1. Therefore, $k$ is $(e^y-1)^-1$. Similarly for $t$.
edited Jul 16 at 0:14
Graham Kemp
80.1k43275
answered Jul 15 at 23:46
NicNic8
3,7113922
@Graham Kemp oops. Thanks. :)
– NicNic8
Jul 16 at 0:24
add a comment |Â
up vote
2
down vote
The integral of a pdf from $-infty$ to $infty$ (the total area under the curve) must be 1. Therefore, $k$ is $(e^y-1)^-1$. Similarly for $t$.
edited Jul 16 at 0:14
Graham Kemp
80.1k43275
answered Jul 15 at 23:46
NicNic8
3,7113922
@Graham Kemp oops. Thanks. :)
– NicNic8
Jul 16 at 0:24
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The integral of a pdf from $-infty$ to $infty$ (the total area under the curve) must be 1. Therefore, $k$ is $(e^y-1)^-1$. Similarly for $t$.
edited Jul 16 at 0:14
Graham Kemp
80.1k43275
answered Jul 15 at 23:46
NicNic8
3,7113922
The integral of a pdf from $-infty$ to $infty$ (the total area under the curve) must be 1. Therefore, $k$ is $(e^y-1)^-1$. Similarly for $t$.
edited Jul 16 at 0:14
Graham Kemp
80.1k43275
answered Jul 15 at 23:46
NicNic8
3,7113922
edited Jul 16 at 0:14
Graham Kemp
80.1k43275
edited Jul 16 at 0:14
Graham Kemp
80.1k43275
edited Jul 16 at 0:14
Graham Kemp
80.1k43275
80.1k43275
answered Jul 15 at 23:46
NicNic8
3,7113922
answered Jul 15 at 23:46
NicNic8
3,7113922
answered Jul 15 at 23:46
NicNic8
3,7113922
3,7113922
@Graham Kemp oops. Thanks. :)
– NicNic8
Jul 16 at 0:24
add a comment |Â
@Graham Kemp oops. Thanks. :)
– NicNic8
Jul 16 at 0:24
@Graham Kemp oops. Thanks. :)
– NicNic8
Jul 16 at 0:24
@Graham Kemp oops. Thanks. :)
– NicNic8
Jul 16 at 0:24
add a comment |Â
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