Mixing
Mean vector and covariance matrix
$begingroup$
I am given a home work for one subject, but my probability theory course is just started, so I dont have enough information. Could someone help me with that?
Given:
$$begin{equation}
p_underline x(x)=left{
begin{array}{@{}ll@{}}
frac1pi, & text{if} x^2_1+x^2_2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation} $$
Find the mean and covariance matrix of the random vector of:
$$ underline y=begin{bmatrix}
1 & -1 \
0 & 2 \
end{bmatrix}underline x + begin{bmatrix} 2 \ 3 \ end{bmatrix}
$$
Marginal distribution, mean and variance is already determined.
Help me please with doing mean and covariance matrix. If this will be explained and possibly given a link to some resource it would be quite helpful for me.
Thanx
matrices probability-theory covariance means
asked Jan 11 at 13:09
Hillbilly JoeHillbilly Joe
164
$endgroup$
add a comment |
$begingroup$
I am given a home work for one subject, but my probability theory course is just started, so I dont have enough information. Could someone help me with that?
Given:
$$begin{equation}
p_underline x(x)=left{
begin{array}{@{}ll@{}}
frac1pi, & text{if} x^2_1+x^2_2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation} $$
Find the mean and covariance matrix of the random vector of:
$$ underline y=begin{bmatrix}
1 & -1 \
0 & 2 \
end{bmatrix}underline x + begin{bmatrix} 2 \ 3 \ end{bmatrix}
$$
Marginal distribution, mean and variance is already determined.
Help me please with doing mean and covariance matrix. If this will be explained and possibly given a link to some resource it would be quite helpful for me.
Thanx
matrices probability-theory covariance means
asked Jan 11 at 13:09
Hillbilly JoeHillbilly Joe
164
$endgroup$
add a comment |
$begingroup$
I am given a home work for one subject, but my probability theory course is just started, so I dont have enough information. Could someone help me with that?
Given:
$$begin{equation}
p_underline x(x)=left{
begin{array}{@{}ll@{}}
frac1pi, & text{if} x^2_1+x^2_2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation} $$
Find the mean and covariance matrix of the random vector of:
$$ underline y=begin{bmatrix}
1 & -1 \
0 & 2 \
end{bmatrix}underline x + begin{bmatrix} 2 \ 3 \ end{bmatrix}
$$
Marginal distribution, mean and variance is already determined.
Help me please with doing mean and covariance matrix. If this will be explained and possibly given a link to some resource it would be quite helpful for me.
Thanx
matrices probability-theory covariance means
asked Jan 11 at 13:09
Hillbilly JoeHillbilly Joe
164
$endgroup$
I am given a home work for one subject, but my probability theory course is just started, so I dont have enough information. Could someone help me with that?
Given:
$$begin{equation}
p_underline x(x)=left{
begin{array}{@{}ll@{}}
frac1pi, & text{if} x^2_1+x^2_2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation} $$
Find the mean and covariance matrix of the random vector of:
$$ underline y=begin{bmatrix}
1 & -1 \
0 & 2 \
end{bmatrix}underline x + begin{bmatrix} 2 \ 3 \ end{bmatrix}
$$
Marginal distribution, mean and variance is already determined.
Help me please with doing mean and covariance matrix. If this will be explained and possibly given a link to some resource it would be quite helpful for me.
Thanx
matrices probability-theory covariance means
matrices probability-theory covariance means
asked Jan 11 at 13:09
Hillbilly JoeHillbilly Joe
164
asked Jan 11 at 13:09
Hillbilly JoeHillbilly Joe
164
asked Jan 11 at 13:09
Hillbilly JoeHillbilly Joe
164
asked Jan 11 at 13:09
Hillbilly JoeHillbilly Joe
164
asked Jan 11 at 13:09
Hillbilly JoeHillbilly Joe
164
164
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: not a complete solution, since you've already done most of the work, given what you've stated. Write
$$underline{y} = begin{bmatrix}
y_1 \
y_2
end{bmatrix}, underline{x} = begin{bmatrix}
x_1 \
x_2
end{bmatrix}text{.}$$
Then we have
$$begin{bmatrix}
y_1 \
y_2
end{bmatrix} = begin{bmatrix}
1 & -1 \
0 & 2
end{bmatrix}begin{bmatrix}
x_1 \
x_2
end{bmatrix} + begin{bmatrix}
2 \
3
end{bmatrix} = begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}text{.}$$
So, what this problem boils down to is finding the mean and covariance matrix of
$$begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}text{.}$$
The expected value is simply
$$mathbb{E}left[begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix} right] = begin{bmatrix}
mathbb{E}[x_1 - x_2 + 2]\
mathbb{E}[2x_2+3]
end{bmatrix}$$
and the covariance matrix is
$$text{Cov}left(begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}right) = begin{bmatrix}
text{Cov}left(x_1 - x_2 + 2, x_1 - x_2 + 2right) & text{Cov}left(x_1 - x_2 + 2, 2x_2+3right) \
text{Cov}left(2x_2+3, x_1 - x_2 + 2right) & text{Cov}left(2x_2+3, 2x_2+3right)
end{bmatrix}text{.}$$
I will let you handle it from here.
answered Jan 11 at 13:24
ClarinetistClarinetist
11k42778
$endgroup$
$begingroup$
Thank you for your help
$endgroup$
– Hillbilly Joe
Jan 11 at 13:29
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint: not a complete solution, since you've already done most of the work, given what you've stated. Write
$$underline{y} = begin{bmatrix}
y_1 \
y_2
end{bmatrix}, underline{x} = begin{bmatrix}
x_1 \
x_2
end{bmatrix}text{.}$$
Then we have
$$begin{bmatrix}
y_1 \
y_2
end{bmatrix} = begin{bmatrix}
1 & -1 \
0 & 2
end{bmatrix}begin{bmatrix}
x_1 \
x_2
end{bmatrix} + begin{bmatrix}
2 \
3
end{bmatrix} = begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}text{.}$$
So, what this problem boils down to is finding the mean and covariance matrix of
$$begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}text{.}$$
The expected value is simply
$$mathbb{E}left[begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix} right] = begin{bmatrix}
mathbb{E}[x_1 - x_2 + 2]\
mathbb{E}[2x_2+3]
end{bmatrix}$$
and the covariance matrix is
$$text{Cov}left(begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}right) = begin{bmatrix}
text{Cov}left(x_1 - x_2 + 2, x_1 - x_2 + 2right) & text{Cov}left(x_1 - x_2 + 2, 2x_2+3right) \
text{Cov}left(2x_2+3, x_1 - x_2 + 2right) & text{Cov}left(2x_2+3, 2x_2+3right)
end{bmatrix}text{.}$$
I will let you handle it from here.
answered Jan 11 at 13:24
ClarinetistClarinetist
11k42778
$endgroup$
$begingroup$
Thank you for your help
$endgroup$
– Hillbilly Joe
Jan 11 at 13:29
add a comment |
$begingroup$
Hint: not a complete solution, since you've already done most of the work, given what you've stated. Write
$$underline{y} = begin{bmatrix}
y_1 \
y_2
end{bmatrix}, underline{x} = begin{bmatrix}
x_1 \
x_2
end{bmatrix}text{.}$$
Then we have
$$begin{bmatrix}
y_1 \
y_2
end{bmatrix} = begin{bmatrix}
1 & -1 \
0 & 2
end{bmatrix}begin{bmatrix}
x_1 \
x_2
end{bmatrix} + begin{bmatrix}
2 \
3
end{bmatrix} = begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}text{.}$$
So, what this problem boils down to is finding the mean and covariance matrix of
$$begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}text{.}$$
The expected value is simply
$$mathbb{E}left[begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix} right] = begin{bmatrix}
mathbb{E}[x_1 - x_2 + 2]\
mathbb{E}[2x_2+3]
end{bmatrix}$$
and the covariance matrix is
$$text{Cov}left(begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}right) = begin{bmatrix}
text{Cov}left(x_1 - x_2 + 2, x_1 - x_2 + 2right) & text{Cov}left(x_1 - x_2 + 2, 2x_2+3right) \
text{Cov}left(2x_2+3, x_1 - x_2 + 2right) & text{Cov}left(2x_2+3, 2x_2+3right)
end{bmatrix}text{.}$$
I will let you handle it from here.
answered Jan 11 at 13:24
ClarinetistClarinetist
11k42778
$endgroup$
$begingroup$
Thank you for your help
$endgroup$
– Hillbilly Joe
Jan 11 at 13:29
add a comment |
$begingroup$
Hint: not a complete solution, since you've already done most of the work, given what you've stated. Write
$$underline{y} = begin{bmatrix}
y_1 \
y_2
end{bmatrix}, underline{x} = begin{bmatrix}
x_1 \
x_2
end{bmatrix}text{.}$$
Then we have
$$begin{bmatrix}
y_1 \
y_2
end{bmatrix} = begin{bmatrix}
1 & -1 \
0 & 2
end{bmatrix}begin{bmatrix}
x_1 \
x_2
end{bmatrix} + begin{bmatrix}
2 \
3
end{bmatrix} = begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}text{.}$$
So, what this problem boils down to is finding the mean and covariance matrix of
$$begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}text{.}$$
The expected value is simply
$$mathbb{E}left[begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix} right] = begin{bmatrix}
mathbb{E}[x_1 - x_2 + 2]\
mathbb{E}[2x_2+3]
end{bmatrix}$$
and the covariance matrix is
$$text{Cov}left(begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}right) = begin{bmatrix}
text{Cov}left(x_1 - x_2 + 2, x_1 - x_2 + 2right) & text{Cov}left(x_1 - x_2 + 2, 2x_2+3right) \
text{Cov}left(2x_2+3, x_1 - x_2 + 2right) & text{Cov}left(2x_2+3, 2x_2+3right)
end{bmatrix}text{.}$$
I will let you handle it from here.
answered Jan 11 at 13:24
ClarinetistClarinetist
11k42778
$endgroup$
Hint: not a complete solution, since you've already done most of the work, given what you've stated. Write
$$underline{y} = begin{bmatrix}
y_1 \
y_2
end{bmatrix}, underline{x} = begin{bmatrix}
x_1 \
x_2
end{bmatrix}text{.}$$
Then we have
$$begin{bmatrix}
y_1 \
y_2
end{bmatrix} = begin{bmatrix}
1 & -1 \
0 & 2
end{bmatrix}begin{bmatrix}
x_1 \
x_2
end{bmatrix} + begin{bmatrix}
2 \
3
end{bmatrix} = begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}text{.}$$
So, what this problem boils down to is finding the mean and covariance matrix of
$$begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}text{.}$$
The expected value is simply
$$mathbb{E}left[begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix} right] = begin{bmatrix}
mathbb{E}[x_1 - x_2 + 2]\
mathbb{E}[2x_2+3]
end{bmatrix}$$
and the covariance matrix is
$$text{Cov}left(begin{bmatrix}
x_1 - x_2 + 2\
2x_2+3
end{bmatrix}right) = begin{bmatrix}
text{Cov}left(x_1 - x_2 + 2, x_1 - x_2 + 2right) & text{Cov}left(x_1 - x_2 + 2, 2x_2+3right) \
text{Cov}left(2x_2+3, x_1 - x_2 + 2right) & text{Cov}left(2x_2+3, 2x_2+3right)
end{bmatrix}text{.}$$
I will let you handle it from here.
answered Jan 11 at 13:24
ClarinetistClarinetist
11k42778
answered Jan 11 at 13:24
ClarinetistClarinetist
11k42778
answered Jan 11 at 13:24
ClarinetistClarinetist
11k42778
answered Jan 11 at 13:24
ClarinetistClarinetist
11k42778
11k42778
$begingroup$
Thank you for your help
$endgroup$
– Hillbilly Joe
Jan 11 at 13:29
add a comment |
$begingroup$
Thank you for your help
$endgroup$
– Hillbilly Joe
Jan 11 at 13:29
$begingroup$
Thank you for your help
$endgroup$
– Hillbilly Joe
Jan 11 at 13:29
$begingroup$
Thank you for your help
$endgroup$
– Hillbilly Joe
Jan 11 at 13:29
add a comment |
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