Suppose that $f (x, y) = xe^{−x(y+1)}$, where $0 ≤ x < ∞$, $0 ≤ y < ∞$. Find marginal...
$begingroup$
This question comes from rice 3.14
Suppose that $$f (x, y) = xe^{−x(y+1)}$$ where $0 ≤ x < ∞$, $0 ≤ y <
> ∞$
a. Find the marginal densities of X and Y . Are X and Y independent?
b. Find the conditional densities of X and Y
to find the marginal densities i have integrated out $x$ and $y$ such that:
begin{align}
f_X(x) & = int_0^infty xe^{−x(y+1)} dx \
& = x int_0^infty e^{−x(y+1)} dx\
& = x Big[ frac{e^{−x(y+2)}}{y+2} Big]_0^infty\
& = -frac{x}{y+2} space text{for} 0leq x < infty
end{align}
begin{align}
f_Y(y) & = int_0^infty xe^{−x(y+1)} dy \
end{align}
but i get stuck here. How to solve the integral? is the first marginal distribution correct?
statistics probability-distributions definite-integrals exponential-distribution
edited Dec 13 '18 at 9:51
TZakrevskiy
20.2k12354
asked Dec 12 '18 at 9:21
user1607user1607
1718
$endgroup$
add a comment |
$begingroup$
This question comes from rice 3.14
Suppose that $$f (x, y) = xe^{−x(y+1)}$$ where $0 ≤ x < ∞$, $0 ≤ y <
> ∞$
a. Find the marginal densities of X and Y . Are X and Y independent?
b. Find the conditional densities of X and Y
to find the marginal densities i have integrated out $x$ and $y$ such that:
begin{align}
f_X(x) & = int_0^infty xe^{−x(y+1)} dx \
& = x int_0^infty e^{−x(y+1)} dx\
& = x Big[ frac{e^{−x(y+2)}}{y+2} Big]_0^infty\
& = -frac{x}{y+2} space text{for} 0leq x < infty
end{align}
begin{align}
f_Y(y) & = int_0^infty xe^{−x(y+1)} dy \
end{align}
but i get stuck here. How to solve the integral? is the first marginal distribution correct?
statistics probability-distributions definite-integrals exponential-distribution
edited Dec 13 '18 at 9:51
TZakrevskiy
20.2k12354
asked Dec 12 '18 at 9:21
user1607user1607
1718
$endgroup$
add a comment |
$begingroup$
This question comes from rice 3.14
Suppose that $$f (x, y) = xe^{−x(y+1)}$$ where $0 ≤ x < ∞$, $0 ≤ y <
> ∞$
a. Find the marginal densities of X and Y . Are X and Y independent?
b. Find the conditional densities of X and Y
to find the marginal densities i have integrated out $x$ and $y$ such that:
begin{align}
f_X(x) & = int_0^infty xe^{−x(y+1)} dx \
& = x int_0^infty e^{−x(y+1)} dx\
& = x Big[ frac{e^{−x(y+2)}}{y+2} Big]_0^infty\
& = -frac{x}{y+2} space text{for} 0leq x < infty
end{align}
begin{align}
f_Y(y) & = int_0^infty xe^{−x(y+1)} dy \
end{align}
but i get stuck here. How to solve the integral? is the first marginal distribution correct?
statistics probability-distributions definite-integrals exponential-distribution
edited Dec 13 '18 at 9:51
TZakrevskiy
20.2k12354
asked Dec 12 '18 at 9:21
user1607user1607
1718
$endgroup$
This question comes from rice 3.14
Suppose that $$f (x, y) = xe^{−x(y+1)}$$ where $0 ≤ x < ∞$, $0 ≤ y <
> ∞$
a. Find the marginal densities of X and Y . Are X and Y independent?
b. Find the conditional densities of X and Y
to find the marginal densities i have integrated out $x$ and $y$ such that:
begin{align}
f_X(x) & = int_0^infty xe^{−x(y+1)} dx \
& = x int_0^infty e^{−x(y+1)} dx\
& = x Big[ frac{e^{−x(y+2)}}{y+2} Big]_0^infty\
& = -frac{x}{y+2} space text{for} 0leq x < infty
end{align}
begin{align}
f_Y(y) & = int_0^infty xe^{−x(y+1)} dy \
end{align}
but i get stuck here. How to solve the integral? is the first marginal distribution correct?
statistics probability-distributions definite-integrals exponential-distribution
statistics probability-distributions definite-integrals exponential-distribution
edited Dec 13 '18 at 9:51
TZakrevskiy
20.2k12354
asked Dec 12 '18 at 9:21
user1607user1607
1718
edited Dec 13 '18 at 9:51
TZakrevskiy
20.2k12354
asked Dec 12 '18 at 9:21
user1607user1607
1718
edited Dec 13 '18 at 9:51
TZakrevskiy
20.2k12354
edited Dec 13 '18 at 9:51
TZakrevskiy
20.2k12354
edited Dec 13 '18 at 9:51
TZakrevskiy
20.2k12354
20.2k12354
asked Dec 12 '18 at 9:21
user1607user1607
1718
asked Dec 12 '18 at 9:21
user1607user1607
1718
asked Dec 12 '18 at 9:21
user1607user1607
1718
1718
add a comment |
add a comment |
1 Answer
1
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$begingroup$
$$f_Y (y)=int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$ $f_X(x)$ is easier: $f_X (x)=int_0^{infty} xe^{-x(y+1)}dy=xe^{-x} int_0^{infty} e^{-xy} dy=e^{-x}$
answered Dec 12 '18 at 9:24
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
$endgroup$
$begingroup$
could you break down the steps of how you obtaine the $f_Y(y)$. I have tries integration per partes twice, but i must be doing a mistake somewhere
$endgroup$
– user1607
Dec 12 '18 at 18:10
add a comment |
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1 Answer
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$begingroup$
$$f_Y (y)=int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$ $f_X(x)$ is easier: $f_X (x)=int_0^{infty} xe^{-x(y+1)}dy=xe^{-x} int_0^{infty} e^{-xy} dy=e^{-x}$
answered Dec 12 '18 at 9:24
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
$endgroup$
$begingroup$
could you break down the steps of how you obtaine the $f_Y(y)$. I have tries integration per partes twice, but i must be doing a mistake somewhere
$endgroup$
– user1607
Dec 12 '18 at 18:10
add a comment |
$begingroup$
$$f_Y (y)=int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$ $f_X(x)$ is easier: $f_X (x)=int_0^{infty} xe^{-x(y+1)}dy=xe^{-x} int_0^{infty} e^{-xy} dy=e^{-x}$
answered Dec 12 '18 at 9:24
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
$endgroup$
$begingroup$
could you break down the steps of how you obtaine the $f_Y(y)$. I have tries integration per partes twice, but i must be doing a mistake somewhere
$endgroup$
– user1607
Dec 12 '18 at 18:10
add a comment |
$begingroup$
$$f_Y (y)=int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$ $f_X(x)$ is easier: $f_X (x)=int_0^{infty} xe^{-x(y+1)}dy=xe^{-x} int_0^{infty} e^{-xy} dy=e^{-x}$
answered Dec 12 '18 at 9:24
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
$endgroup$
$$f_Y (y)=int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$ $f_X(x)$ is easier: $f_X (x)=int_0^{infty} xe^{-x(y+1)}dy=xe^{-x} int_0^{infty} e^{-xy} dy=e^{-x}$
answered Dec 12 '18 at 9:24
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
edited Dec 12 '18 at 9:30
answered Dec 12 '18 at 9:24
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
answered Dec 12 '18 at 9:24
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
answered Dec 12 '18 at 9:24
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
71.7k53170
$begingroup$
could you break down the steps of how you obtaine the $f_Y(y)$. I have tries integration per partes twice, but i must be doing a mistake somewhere
$endgroup$
– user1607
Dec 12 '18 at 18:10
add a comment |
$begingroup$
could you break down the steps of how you obtaine the $f_Y(y)$. I have tries integration per partes twice, but i must be doing a mistake somewhere
$endgroup$
– user1607
Dec 12 '18 at 18:10
$begingroup$
could you break down the steps of how you obtaine the $f_Y(y)$. I have tries integration per partes twice, but i must be doing a mistake somewhere
$endgroup$
– user1607
Dec 12 '18 at 18:10
$begingroup$
could you break down the steps of how you obtaine the $f_Y(y)$. I have tries integration per partes twice, but i must be doing a mistake somewhere
$endgroup$
– user1607
Dec 12 '18 at 18:10
add a comment |
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