Showing that $X + Y$ and $X/Y$ are independent given that $X$ and $Y$ are i.i.d with a PDF
Let $X$ and $Y$ be i.i.d with pdfs given by
$$f_{X}(x) = begin{cases}
text{exp}(-x), & text{ if } x > 0, \
0 & text{ otherwise},
end{cases} $$
and
$$f_{Y}(y) = begin{cases}
text{exp}(-y), & text{ if } x > 0, \
0 & text{ otherwise}.
end{cases} $$
Show that $X + Y$ and $X/Y$ are independent.
My attempt:
By independence, we have
$$f_{X, Y}(x, y) = begin{cases}
text{exp}(-(x + y)) & text{ if } x+y > 0 \
0 & text{ otherwise.}
end{cases} $$
Let $Y_{1} = g_{1}(X, Y) = X + Y$ and $Y_{2} = g_{2}(X, Y) = X/Y$. Solving for $X$ and $Y$ gives us the solutions $X = Y_{1}Y_{2}/(Y_{2} + 1)$ and $Y = Y_{1}/(Y_{2} + 1)$. Thus, our Jacobian is given by
$$J = frac{partial g_{1}}{partial X}frac{partial g_{2}}{partial Y} - frac{partial g_{1}}{partial Y}frac{partial g_{2}}{partial X} $$
$$= -frac{X + Y}{Y^{2}} $$
In terms of $Y_{1}$ and $Y_{2}$, we have
$$-frac{X + Y}{Y^{2}} = -frac{(Y_{2} + 1)^{2}}{Y_{1}}.$$
Therefore, our new probability density function is given by
$$f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) = f(x_{1}, x_{2}) cdot |J|^{-1} $$
$$= begin{cases}
dfrac{y_{1}}{(y_{2} + 1)^{2}} cdot text{exp}(-y_{1}) & text{ if } y_{1} > 0 \[1em]
0 & text{ otherwise.}
end{cases} $$
So, we compute the marginal densities as follows:
$$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{-infty}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}},$$
but the integral on the right diverges. Where did I go wrong? Can't figure it out. Also, did I do anything else wrong (in particular, are my domains for $f_{X, Y}$ and $f_{Y_{1}, Y_{2}}$ correct?)
probability probability-theory
asked Nov 24 at 4:14
stackofhay42
1696
add a comment |
Let $X$ and $Y$ be i.i.d with pdfs given by
$$f_{X}(x) = begin{cases}
text{exp}(-x), & text{ if } x > 0, \
0 & text{ otherwise},
end{cases} $$
and
$$f_{Y}(y) = begin{cases}
text{exp}(-y), & text{ if } x > 0, \
0 & text{ otherwise}.
end{cases} $$
Show that $X + Y$ and $X/Y$ are independent.
My attempt:
By independence, we have
$$f_{X, Y}(x, y) = begin{cases}
text{exp}(-(x + y)) & text{ if } x+y > 0 \
0 & text{ otherwise.}
end{cases} $$
Let $Y_{1} = g_{1}(X, Y) = X + Y$ and $Y_{2} = g_{2}(X, Y) = X/Y$. Solving for $X$ and $Y$ gives us the solutions $X = Y_{1}Y_{2}/(Y_{2} + 1)$ and $Y = Y_{1}/(Y_{2} + 1)$. Thus, our Jacobian is given by
$$J = frac{partial g_{1}}{partial X}frac{partial g_{2}}{partial Y} - frac{partial g_{1}}{partial Y}frac{partial g_{2}}{partial X} $$
$$= -frac{X + Y}{Y^{2}} $$
In terms of $Y_{1}$ and $Y_{2}$, we have
$$-frac{X + Y}{Y^{2}} = -frac{(Y_{2} + 1)^{2}}{Y_{1}}.$$
Therefore, our new probability density function is given by
$$f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) = f(x_{1}, x_{2}) cdot |J|^{-1} $$
$$= begin{cases}
dfrac{y_{1}}{(y_{2} + 1)^{2}} cdot text{exp}(-y_{1}) & text{ if } y_{1} > 0 \[1em]
0 & text{ otherwise.}
end{cases} $$
So, we compute the marginal densities as follows:
$$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{-infty}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}},$$
but the integral on the right diverges. Where did I go wrong? Can't figure it out. Also, did I do anything else wrong (in particular, are my domains for $f_{X, Y}$ and $f_{Y_{1}, Y_{2}}$ correct?)
probability probability-theory
asked Nov 24 at 4:14
stackofhay42
1696
add a comment |
Let $X$ and $Y$ be i.i.d with pdfs given by
$$f_{X}(x) = begin{cases}
text{exp}(-x), & text{ if } x > 0, \
0 & text{ otherwise},
end{cases} $$
and
$$f_{Y}(y) = begin{cases}
text{exp}(-y), & text{ if } x > 0, \
0 & text{ otherwise}.
end{cases} $$
Show that $X + Y$ and $X/Y$ are independent.
My attempt:
By independence, we have
$$f_{X, Y}(x, y) = begin{cases}
text{exp}(-(x + y)) & text{ if } x+y > 0 \
0 & text{ otherwise.}
end{cases} $$
Let $Y_{1} = g_{1}(X, Y) = X + Y$ and $Y_{2} = g_{2}(X, Y) = X/Y$. Solving for $X$ and $Y$ gives us the solutions $X = Y_{1}Y_{2}/(Y_{2} + 1)$ and $Y = Y_{1}/(Y_{2} + 1)$. Thus, our Jacobian is given by
$$J = frac{partial g_{1}}{partial X}frac{partial g_{2}}{partial Y} - frac{partial g_{1}}{partial Y}frac{partial g_{2}}{partial X} $$
$$= -frac{X + Y}{Y^{2}} $$
In terms of $Y_{1}$ and $Y_{2}$, we have
$$-frac{X + Y}{Y^{2}} = -frac{(Y_{2} + 1)^{2}}{Y_{1}}.$$
Therefore, our new probability density function is given by
$$f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) = f(x_{1}, x_{2}) cdot |J|^{-1} $$
$$= begin{cases}
dfrac{y_{1}}{(y_{2} + 1)^{2}} cdot text{exp}(-y_{1}) & text{ if } y_{1} > 0 \[1em]
0 & text{ otherwise.}
end{cases} $$
So, we compute the marginal densities as follows:
$$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{-infty}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}},$$
but the integral on the right diverges. Where did I go wrong? Can't figure it out. Also, did I do anything else wrong (in particular, are my domains for $f_{X, Y}$ and $f_{Y_{1}, Y_{2}}$ correct?)
probability probability-theory
asked Nov 24 at 4:14
stackofhay42
1696
Let $X$ and $Y$ be i.i.d with pdfs given by
$$f_{X}(x) = begin{cases}
text{exp}(-x), & text{ if } x > 0, \
0 & text{ otherwise},
end{cases} $$
and
$$f_{Y}(y) = begin{cases}
text{exp}(-y), & text{ if } x > 0, \
0 & text{ otherwise}.
end{cases} $$
Show that $X + Y$ and $X/Y$ are independent.
My attempt:
By independence, we have
$$f_{X, Y}(x, y) = begin{cases}
text{exp}(-(x + y)) & text{ if } x+y > 0 \
0 & text{ otherwise.}
end{cases} $$
Let $Y_{1} = g_{1}(X, Y) = X + Y$ and $Y_{2} = g_{2}(X, Y) = X/Y$. Solving for $X$ and $Y$ gives us the solutions $X = Y_{1}Y_{2}/(Y_{2} + 1)$ and $Y = Y_{1}/(Y_{2} + 1)$. Thus, our Jacobian is given by
$$J = frac{partial g_{1}}{partial X}frac{partial g_{2}}{partial Y} - frac{partial g_{1}}{partial Y}frac{partial g_{2}}{partial X} $$
$$= -frac{X + Y}{Y^{2}} $$
In terms of $Y_{1}$ and $Y_{2}$, we have
$$-frac{X + Y}{Y^{2}} = -frac{(Y_{2} + 1)^{2}}{Y_{1}}.$$
Therefore, our new probability density function is given by
$$f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) = f(x_{1}, x_{2}) cdot |J|^{-1} $$
$$= begin{cases}
dfrac{y_{1}}{(y_{2} + 1)^{2}} cdot text{exp}(-y_{1}) & text{ if } y_{1} > 0 \[1em]
0 & text{ otherwise.}
end{cases} $$
So, we compute the marginal densities as follows:
$$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{-infty}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}},$$
but the integral on the right diverges. Where did I go wrong? Can't figure it out. Also, did I do anything else wrong (in particular, are my domains for $f_{X, Y}$ and $f_{Y_{1}, Y_{2}}$ correct?)
probability probability-theory
probability probability-theory
asked Nov 24 at 4:14
stackofhay42
1696
asked Nov 24 at 4:14
stackofhay42
1696
asked Nov 24 at 4:14
stackofhay42
1696
asked Nov 24 at 4:14
stackofhay42
1696
asked Nov 24 at 4:14
stackofhay42
1696
1696
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
We have if $y_1 > 0$, then
$$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{color{red}0}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}} = y_1 exp(-y_1)$$
answered Nov 24 at 4:22
Siong Thye Goh
98.9k1464116
Why is the lower bound $0$?
– stackofhay42
Nov 24 at 4:26
$y_2$ only take positive values right since it is a division of two positive numbers.
– Siong Thye Goh
Nov 24 at 4:27
Ok. Then, $y_{1}$ must only take positive values too, right?
– stackofhay42
Nov 24 at 4:28
yes, addition of positive numbers.
– Siong Thye Goh
Nov 24 at 4:29
add a comment |
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1 Answer
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We have if $y_1 > 0$, then
$$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{color{red}0}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}} = y_1 exp(-y_1)$$
answered Nov 24 at 4:22
Siong Thye Goh
98.9k1464116
Why is the lower bound $0$?
– stackofhay42
Nov 24 at 4:26
$y_2$ only take positive values right since it is a division of two positive numbers.
– Siong Thye Goh
Nov 24 at 4:27
Ok. Then, $y_{1}$ must only take positive values too, right?
– stackofhay42
Nov 24 at 4:28
yes, addition of positive numbers.
– Siong Thye Goh
Nov 24 at 4:29
add a comment |
We have if $y_1 > 0$, then
$$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{color{red}0}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}} = y_1 exp(-y_1)$$
answered Nov 24 at 4:22
Siong Thye Goh
98.9k1464116
Why is the lower bound $0$?
– stackofhay42
Nov 24 at 4:26
$y_2$ only take positive values right since it is a division of two positive numbers.
– Siong Thye Goh
Nov 24 at 4:27
Ok. Then, $y_{1}$ must only take positive values too, right?
– stackofhay42
Nov 24 at 4:28
yes, addition of positive numbers.
– Siong Thye Goh
Nov 24 at 4:29
add a comment |
We have if $y_1 > 0$, then
$$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{color{red}0}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}} = y_1 exp(-y_1)$$
answered Nov 24 at 4:22
Siong Thye Goh
98.9k1464116
We have if $y_1 > 0$, then
$$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{color{red}0}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}} = y_1 exp(-y_1)$$
answered Nov 24 at 4:22
Siong Thye Goh
98.9k1464116
answered Nov 24 at 4:22
Siong Thye Goh
98.9k1464116
answered Nov 24 at 4:22
Siong Thye Goh
98.9k1464116
answered Nov 24 at 4:22
Siong Thye Goh
98.9k1464116
98.9k1464116
Why is the lower bound $0$?
– stackofhay42
Nov 24 at 4:26
$y_2$ only take positive values right since it is a division of two positive numbers.
– Siong Thye Goh
Nov 24 at 4:27
Ok. Then, $y_{1}$ must only take positive values too, right?
– stackofhay42
Nov 24 at 4:28
yes, addition of positive numbers.
– Siong Thye Goh
Nov 24 at 4:29
add a comment |
Why is the lower bound $0$?
– stackofhay42
Nov 24 at 4:26
$y_2$ only take positive values right since it is a division of two positive numbers.
– Siong Thye Goh
Nov 24 at 4:27
Ok. Then, $y_{1}$ must only take positive values too, right?
– stackofhay42
Nov 24 at 4:28
yes, addition of positive numbers.
– Siong Thye Goh
Nov 24 at 4:29
Why is the lower bound $0$?
– stackofhay42
Nov 24 at 4:26
Why is the lower bound $0$?
– stackofhay42
Nov 24 at 4:26
$y_2$ only take positive values right since it is a division of two positive numbers.
– Siong Thye Goh
Nov 24 at 4:27
$y_2$ only take positive values right since it is a division of two positive numbers.
– Siong Thye Goh
Nov 24 at 4:27
Ok. Then, $y_{1}$ must only take positive values too, right?
– stackofhay42
Nov 24 at 4:28
Ok. Then, $y_{1}$ must only take positive values too, right?
– stackofhay42
Nov 24 at 4:28
yes, addition of positive numbers.
– Siong Thye Goh
Nov 24 at 4:29
yes, addition of positive numbers.
– Siong Thye Goh
Nov 24 at 4:29
add a comment |
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