joint PDF of 2 dependent variables w/ convolution
$begingroup$
Let $X, Y$ be iid standard uniform variables, and let $T = X + Y$. The goal is to find the joint PDF of $X$ and $T$. The work I've done so far is to find the PDF of $T$ by evaluating the convolution, but I'm not even sure if that's useful because $X$ and $T$ aren't independent, so the joint PDF shouldn't be the product of the marginal PDFs. I thought about working from the CDF and writing it as $P(X leq x, T leq t) = P(X leq x, X + Y leq t)$. I didn't know where to go from there due to the lack of independence. Can anyone provide suggestions?
probability probability-distributions random-variables uniform-distribution
asked Dec 10 '18 at 22:37
0k330k33
12210
$endgroup$
add a comment |
$begingroup$
Let $X, Y$ be iid standard uniform variables, and let $T = X + Y$. The goal is to find the joint PDF of $X$ and $T$. The work I've done so far is to find the PDF of $T$ by evaluating the convolution, but I'm not even sure if that's useful because $X$ and $T$ aren't independent, so the joint PDF shouldn't be the product of the marginal PDFs. I thought about working from the CDF and writing it as $P(X leq x, T leq t) = P(X leq x, X + Y leq t)$. I didn't know where to go from there due to the lack of independence. Can anyone provide suggestions?
probability probability-distributions random-variables uniform-distribution
asked Dec 10 '18 at 22:37
0k330k33
12210
$endgroup$
add a comment |
$begingroup$
Let $X, Y$ be iid standard uniform variables, and let $T = X + Y$. The goal is to find the joint PDF of $X$ and $T$. The work I've done so far is to find the PDF of $T$ by evaluating the convolution, but I'm not even sure if that's useful because $X$ and $T$ aren't independent, so the joint PDF shouldn't be the product of the marginal PDFs. I thought about working from the CDF and writing it as $P(X leq x, T leq t) = P(X leq x, X + Y leq t)$. I didn't know where to go from there due to the lack of independence. Can anyone provide suggestions?
probability probability-distributions random-variables uniform-distribution
asked Dec 10 '18 at 22:37
0k330k33
12210
$endgroup$
Let $X, Y$ be iid standard uniform variables, and let $T = X + Y$. The goal is to find the joint PDF of $X$ and $T$. The work I've done so far is to find the PDF of $T$ by evaluating the convolution, but I'm not even sure if that's useful because $X$ and $T$ aren't independent, so the joint PDF shouldn't be the product of the marginal PDFs. I thought about working from the CDF and writing it as $P(X leq x, T leq t) = P(X leq x, X + Y leq t)$. I didn't know where to go from there due to the lack of independence. Can anyone provide suggestions?
probability probability-distributions random-variables uniform-distribution
probability probability-distributions random-variables uniform-distribution
asked Dec 10 '18 at 22:37
0k330k33
12210
asked Dec 10 '18 at 22:37
0k330k33
12210
asked Dec 10 '18 at 22:37
0k330k33
12210
asked Dec 10 '18 at 22:37
0k330k33
12210
asked Dec 10 '18 at 22:37
0k330k33
12210
12210
add a comment |
add a comment |
1 Answer
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$begingroup$
You are almost there. Note that
$$
P(Xle x, T le t) = P(X le x, X+Y le t) = P(X le x, X le t - Y),
$$
It follows from
$$
P(X le x, X le t - Y| Y = y) = int_0^{min{x, t- y}} du = min{x, t-y}
$$
that
begin{align*}
&P(X le x, X le t - Y) \
&= int_0^1 P(X le x, X le t - Y| Y = y) dy \
&= int_0^1 min{x, t-y} dy \
&= int_0^{max{t-x,0}} x dy + int_{max{t-x,0}}^1 (t-y)dy \
&= xmax{t-x,0} + (1-max{t-x,0})t -frac{1 - (max{t-x,0})^2}{2}.
end{align*}
answered Dec 10 '18 at 22:48
induction601induction601
1,162314
$endgroup$
$begingroup$
Thank you so much! I definitely follow your explanation. I was wondering if it also might be possible to compute the PDF of X,T via a linear transformation. Since T = X + Y, would it be correct to say that $begin{bmatrix}1&0\1&1end{bmatrix} (X, Y) = (X, T)$? And the matrix has determinant $1$. But then we have that $f_{X,T} = f_{X, Y}(x, t-x}$, which is still $1$ because $f_{X, Y}$ is 1 on the support of $X, Y$. I am sure that something I did is not right?!
$endgroup$
– 0k33
Dec 12 '18 at 21:54
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
You are almost there. Note that
$$
P(Xle x, T le t) = P(X le x, X+Y le t) = P(X le x, X le t - Y),
$$
It follows from
$$
P(X le x, X le t - Y| Y = y) = int_0^{min{x, t- y}} du = min{x, t-y}
$$
that
begin{align*}
&P(X le x, X le t - Y) \
&= int_0^1 P(X le x, X le t - Y| Y = y) dy \
&= int_0^1 min{x, t-y} dy \
&= int_0^{max{t-x,0}} x dy + int_{max{t-x,0}}^1 (t-y)dy \
&= xmax{t-x,0} + (1-max{t-x,0})t -frac{1 - (max{t-x,0})^2}{2}.
end{align*}
answered Dec 10 '18 at 22:48
induction601induction601
1,162314
$endgroup$
$begingroup$
Thank you so much! I definitely follow your explanation. I was wondering if it also might be possible to compute the PDF of X,T via a linear transformation. Since T = X + Y, would it be correct to say that $begin{bmatrix}1&0\1&1end{bmatrix} (X, Y) = (X, T)$? And the matrix has determinant $1$. But then we have that $f_{X,T} = f_{X, Y}(x, t-x}$, which is still $1$ because $f_{X, Y}$ is 1 on the support of $X, Y$. I am sure that something I did is not right?!
$endgroup$
– 0k33
Dec 12 '18 at 21:54
add a comment |
$begingroup$
You are almost there. Note that
$$
P(Xle x, T le t) = P(X le x, X+Y le t) = P(X le x, X le t - Y),
$$
It follows from
$$
P(X le x, X le t - Y| Y = y) = int_0^{min{x, t- y}} du = min{x, t-y}
$$
that
begin{align*}
&P(X le x, X le t - Y) \
&= int_0^1 P(X le x, X le t - Y| Y = y) dy \
&= int_0^1 min{x, t-y} dy \
&= int_0^{max{t-x,0}} x dy + int_{max{t-x,0}}^1 (t-y)dy \
&= xmax{t-x,0} + (1-max{t-x,0})t -frac{1 - (max{t-x,0})^2}{2}.
end{align*}
answered Dec 10 '18 at 22:48
induction601induction601
1,162314
$endgroup$
$begingroup$
Thank you so much! I definitely follow your explanation. I was wondering if it also might be possible to compute the PDF of X,T via a linear transformation. Since T = X + Y, would it be correct to say that $begin{bmatrix}1&0\1&1end{bmatrix} (X, Y) = (X, T)$? And the matrix has determinant $1$. But then we have that $f_{X,T} = f_{X, Y}(x, t-x}$, which is still $1$ because $f_{X, Y}$ is 1 on the support of $X, Y$. I am sure that something I did is not right?!
$endgroup$
– 0k33
Dec 12 '18 at 21:54
add a comment |
$begingroup$
You are almost there. Note that
$$
P(Xle x, T le t) = P(X le x, X+Y le t) = P(X le x, X le t - Y),
$$
It follows from
$$
P(X le x, X le t - Y| Y = y) = int_0^{min{x, t- y}} du = min{x, t-y}
$$
that
begin{align*}
&P(X le x, X le t - Y) \
&= int_0^1 P(X le x, X le t - Y| Y = y) dy \
&= int_0^1 min{x, t-y} dy \
&= int_0^{max{t-x,0}} x dy + int_{max{t-x,0}}^1 (t-y)dy \
&= xmax{t-x,0} + (1-max{t-x,0})t -frac{1 - (max{t-x,0})^2}{2}.
end{align*}
answered Dec 10 '18 at 22:48
induction601induction601
1,162314
$endgroup$
You are almost there. Note that
$$
P(Xle x, T le t) = P(X le x, X+Y le t) = P(X le x, X le t - Y),
$$
It follows from
$$
P(X le x, X le t - Y| Y = y) = int_0^{min{x, t- y}} du = min{x, t-y}
$$
that
begin{align*}
&P(X le x, X le t - Y) \
&= int_0^1 P(X le x, X le t - Y| Y = y) dy \
&= int_0^1 min{x, t-y} dy \
&= int_0^{max{t-x,0}} x dy + int_{max{t-x,0}}^1 (t-y)dy \
&= xmax{t-x,0} + (1-max{t-x,0})t -frac{1 - (max{t-x,0})^2}{2}.
end{align*}
answered Dec 10 '18 at 22:48
induction601induction601
1,162314
answered Dec 10 '18 at 22:48
induction601induction601
1,162314
answered Dec 10 '18 at 22:48
induction601induction601
1,162314
answered Dec 10 '18 at 22:48
induction601induction601
1,162314
1,162314
$begingroup$
Thank you so much! I definitely follow your explanation. I was wondering if it also might be possible to compute the PDF of X,T via a linear transformation. Since T = X + Y, would it be correct to say that $begin{bmatrix}1&0\1&1end{bmatrix} (X, Y) = (X, T)$? And the matrix has determinant $1$. But then we have that $f_{X,T} = f_{X, Y}(x, t-x}$, which is still $1$ because $f_{X, Y}$ is 1 on the support of $X, Y$. I am sure that something I did is not right?!
$endgroup$
– 0k33
Dec 12 '18 at 21:54
add a comment |
$begingroup$
Thank you so much! I definitely follow your explanation. I was wondering if it also might be possible to compute the PDF of X,T via a linear transformation. Since T = X + Y, would it be correct to say that $begin{bmatrix}1&0\1&1end{bmatrix} (X, Y) = (X, T)$? And the matrix has determinant $1$. But then we have that $f_{X,T} = f_{X, Y}(x, t-x}$, which is still $1$ because $f_{X, Y}$ is 1 on the support of $X, Y$. I am sure that something I did is not right?!
$endgroup$
– 0k33
Dec 12 '18 at 21:54
$begingroup$
Thank you so much! I definitely follow your explanation. I was wondering if it also might be possible to compute the PDF of X,T via a linear transformation. Since T = X + Y, would it be correct to say that $begin{bmatrix}1&0\1&1end{bmatrix} (X, Y) = (X, T)$? And the matrix has determinant $1$. But then we have that $f_{X,T} = f_{X, Y}(x, t-x}$, which is still $1$ because $f_{X, Y}$ is 1 on the support of $X, Y$. I am sure that something I did is not right?!
$endgroup$
– 0k33
Dec 12 '18 at 21:54
$begingroup$
Thank you so much! I definitely follow your explanation. I was wondering if it also might be possible to compute the PDF of X,T via a linear transformation. Since T = X + Y, would it be correct to say that $begin{bmatrix}1&0\1&1end{bmatrix} (X, Y) = (X, T)$? And the matrix has determinant $1$. But then we have that $f_{X,T} = f_{X, Y}(x, t-x}$, which is still $1$ because $f_{X, Y}$ is 1 on the support of $X, Y$. I am sure that something I did is not right?!
$endgroup$
– 0k33
Dec 12 '18 at 21:54
add a comment |
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