Mixing
How to find sum of random vectors (continuous and discrete)?
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After a search on the web I found this can be done with convolutions, though there was no such a topic in our class. Is there other way to solve this?
probability probability-distributions
asked Jan 30 at 1:53
Murad DavudovMurad Davudov
234
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add a comment |
$begingroup$
After a search on the web I found this can be done with convolutions, though there was no such a topic in our class. Is there other way to solve this?
probability probability-distributions
asked Jan 30 at 1:53
Murad DavudovMurad Davudov
234
$endgroup$
add a comment |
$begingroup$
After a search on the web I found this can be done with convolutions, though there was no such a topic in our class. Is there other way to solve this?
probability probability-distributions
asked Jan 30 at 1:53
Murad DavudovMurad Davudov
234
$endgroup$
After a search on the web I found this can be done with convolutions, though there was no such a topic in our class. Is there other way to solve this?
probability probability-distributions
probability probability-distributions
asked Jan 30 at 1:53
Murad DavudovMurad Davudov
234
asked Jan 30 at 1:53
Murad DavudovMurad Davudov
234
asked Jan 30 at 1:53
Murad DavudovMurad Davudov
234
asked Jan 30 at 1:53
Murad DavudovMurad Davudov
234
asked Jan 30 at 1:53
Murad DavudovMurad Davudov
234
234
add a comment |
add a comment |
1 Answer
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$begingroup$
Any solution is going to be using convolutions, you just don't need to call them that.
For example, in problem 4, $Z$ could be equal to $-1, 1, 2, 3,$ or $4$. For each one of those, you compute the probability $P(Z=z)$ by summing over all the possible cases that would lead to that, i.e. $(X,Y)=(x,y),; x+y = z$. In this case the only nontrivial one is
$$P(Z=1) = P((X,Y) = (0,1)) + P((X,Y) = (2,-1)).$$
Problem 5 is the same principle, but continuous. The CDF of $Z$, $G(z)$, is given by the integral of the given distribution over the region $x+y le z$, which is to say the area of that region intersected with the unit square. The PDF of $Z$ is the derivative of that, which will come out to be the length of the line $x+y=z$ intersected with the square.
answered Jan 30 at 2:57
Nathaniel MayerNathaniel Mayer
1,863516
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$begingroup$
4th problem became clear to me, but I have problem grasping the idea of integral over that region. How is it denoted?
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– Murad Davudov
Jan 30 at 5:28
1
$begingroup$
$G(z) = int_{-infty}^z int_{-infty}^{z-y} f(x,y),dx,dy$
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– Nathaniel Mayer
Jan 30 at 7:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Any solution is going to be using convolutions, you just don't need to call them that.
For example, in problem 4, $Z$ could be equal to $-1, 1, 2, 3,$ or $4$. For each one of those, you compute the probability $P(Z=z)$ by summing over all the possible cases that would lead to that, i.e. $(X,Y)=(x,y),; x+y = z$. In this case the only nontrivial one is
$$P(Z=1) = P((X,Y) = (0,1)) + P((X,Y) = (2,-1)).$$
Problem 5 is the same principle, but continuous. The CDF of $Z$, $G(z)$, is given by the integral of the given distribution over the region $x+y le z$, which is to say the area of that region intersected with the unit square. The PDF of $Z$ is the derivative of that, which will come out to be the length of the line $x+y=z$ intersected with the square.
answered Jan 30 at 2:57
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
4th problem became clear to me, but I have problem grasping the idea of integral over that region. How is it denoted?
$endgroup$
– Murad Davudov
Jan 30 at 5:28
1
$begingroup$
$G(z) = int_{-infty}^z int_{-infty}^{z-y} f(x,y),dx,dy$
$endgroup$
– Nathaniel Mayer
Jan 30 at 7:44
add a comment |
$begingroup$
Any solution is going to be using convolutions, you just don't need to call them that.
For example, in problem 4, $Z$ could be equal to $-1, 1, 2, 3,$ or $4$. For each one of those, you compute the probability $P(Z=z)$ by summing over all the possible cases that would lead to that, i.e. $(X,Y)=(x,y),; x+y = z$. In this case the only nontrivial one is
$$P(Z=1) = P((X,Y) = (0,1)) + P((X,Y) = (2,-1)).$$
Problem 5 is the same principle, but continuous. The CDF of $Z$, $G(z)$, is given by the integral of the given distribution over the region $x+y le z$, which is to say the area of that region intersected with the unit square. The PDF of $Z$ is the derivative of that, which will come out to be the length of the line $x+y=z$ intersected with the square.
answered Jan 30 at 2:57
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
4th problem became clear to me, but I have problem grasping the idea of integral over that region. How is it denoted?
$endgroup$
– Murad Davudov
Jan 30 at 5:28
1
$begingroup$
$G(z) = int_{-infty}^z int_{-infty}^{z-y} f(x,y),dx,dy$
$endgroup$
– Nathaniel Mayer
Jan 30 at 7:44
add a comment |
$begingroup$
Any solution is going to be using convolutions, you just don't need to call them that.
For example, in problem 4, $Z$ could be equal to $-1, 1, 2, 3,$ or $4$. For each one of those, you compute the probability $P(Z=z)$ by summing over all the possible cases that would lead to that, i.e. $(X,Y)=(x,y),; x+y = z$. In this case the only nontrivial one is
$$P(Z=1) = P((X,Y) = (0,1)) + P((X,Y) = (2,-1)).$$
Problem 5 is the same principle, but continuous. The CDF of $Z$, $G(z)$, is given by the integral of the given distribution over the region $x+y le z$, which is to say the area of that region intersected with the unit square. The PDF of $Z$ is the derivative of that, which will come out to be the length of the line $x+y=z$ intersected with the square.
answered Jan 30 at 2:57
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
Any solution is going to be using convolutions, you just don't need to call them that.
For example, in problem 4, $Z$ could be equal to $-1, 1, 2, 3,$ or $4$. For each one of those, you compute the probability $P(Z=z)$ by summing over all the possible cases that would lead to that, i.e. $(X,Y)=(x,y),; x+y = z$. In this case the only nontrivial one is
$$P(Z=1) = P((X,Y) = (0,1)) + P((X,Y) = (2,-1)).$$
Problem 5 is the same principle, but continuous. The CDF of $Z$, $G(z)$, is given by the integral of the given distribution over the region $x+y le z$, which is to say the area of that region intersected with the unit square. The PDF of $Z$ is the derivative of that, which will come out to be the length of the line $x+y=z$ intersected with the square.
answered Jan 30 at 2:57
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 2:57
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 2:57
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 2:57
Nathaniel MayerNathaniel Mayer
1,863516
1,863516
$begingroup$
4th problem became clear to me, but I have problem grasping the idea of integral over that region. How is it denoted?
$endgroup$
– Murad Davudov
Jan 30 at 5:28
1
$begingroup$
$G(z) = int_{-infty}^z int_{-infty}^{z-y} f(x,y),dx,dy$
$endgroup$
– Nathaniel Mayer
Jan 30 at 7:44
add a comment |
$begingroup$
4th problem became clear to me, but I have problem grasping the idea of integral over that region. How is it denoted?
$endgroup$
– Murad Davudov
Jan 30 at 5:28
1
$begingroup$
$G(z) = int_{-infty}^z int_{-infty}^{z-y} f(x,y),dx,dy$
$endgroup$
– Nathaniel Mayer
Jan 30 at 7:44
$begingroup$
4th problem became clear to me, but I have problem grasping the idea of integral over that region. How is it denoted?
$endgroup$
– Murad Davudov
Jan 30 at 5:28
$begingroup$
4th problem became clear to me, but I have problem grasping the idea of integral over that region. How is it denoted?
$endgroup$
– Murad Davudov
Jan 30 at 5:28
1
1
$begingroup$
$G(z) = int_{-infty}^z int_{-infty}^{z-y} f(x,y),dx,dy$
$endgroup$
– Nathaniel Mayer
Jan 30 at 7:44
$begingroup$
$G(z) = int_{-infty}^z int_{-infty}^{z-y} f(x,y),dx,dy$
$endgroup$
– Nathaniel Mayer
Jan 30 at 7:44
add a comment |
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