Deduce X from a constant in variance properties
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How do you consider $operatornameVar(c-alpha X)$?
I'm learning that $V(c) = 0$, where $c$ is a constant.
Does it turn the above into:
$0 - operatornameVar(alpha X) = -[alpha^2operatornameVar(X)]$
statistics variance
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up vote
0
down vote
favorite
How do you consider $operatornameVar(c-alpha X)$?
I'm learning that $V(c) = 0$, where $c$ is a constant.
Does it turn the above into:
$0 - operatornameVar(alpha X) = -[alpha^2operatornameVar(X)]$
statistics variance
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do you consider $operatornameVar(c-alpha X)$?
I'm learning that $V(c) = 0$, where $c$ is a constant.
Does it turn the above into:
$0 - operatornameVar(alpha X) = -[alpha^2operatornameVar(X)]$
statistics variance
How do you consider $operatornameVar(c-alpha X)$?
I'm learning that $V(c) = 0$, where $c$ is a constant.
Does it turn the above into:
$0 - operatornameVar(alpha X) = -[alpha^2operatornameVar(X)]$
statistics variance
statistics variance
edited Sep 7 at 8:54
Bernard
112k635104
112k635104
asked Sep 7 at 8:02
Shimmy
1368
1368
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add a comment |Â
1 Answer
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accepted
beginalign
Var(c-aX) &= Var(-aX) \
&= (-a)^2Var(X)
endalign
We have $$Var(aX+bY) = a^2Var(X)+b^2Var(Y) + 2abCov(X,Y)$$
Thank you very much!
â Shimmy
Sep 7 at 8:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
beginalign
Var(c-aX) &= Var(-aX) \
&= (-a)^2Var(X)
endalign
We have $$Var(aX+bY) = a^2Var(X)+b^2Var(Y) + 2abCov(X,Y)$$
Thank you very much!
â Shimmy
Sep 7 at 8:04
add a comment |Â
up vote
2
down vote
accepted
beginalign
Var(c-aX) &= Var(-aX) \
&= (-a)^2Var(X)
endalign
We have $$Var(aX+bY) = a^2Var(X)+b^2Var(Y) + 2abCov(X,Y)$$
Thank you very much!
â Shimmy
Sep 7 at 8:04
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
beginalign
Var(c-aX) &= Var(-aX) \
&= (-a)^2Var(X)
endalign
We have $$Var(aX+bY) = a^2Var(X)+b^2Var(Y) + 2abCov(X,Y)$$
beginalign
Var(c-aX) &= Var(-aX) \
&= (-a)^2Var(X)
endalign
We have $$Var(aX+bY) = a^2Var(X)+b^2Var(Y) + 2abCov(X,Y)$$
edited Sep 7 at 8:05
answered Sep 7 at 8:04
Siong Thye Goh
82.7k1456104
82.7k1456104
Thank you very much!
â Shimmy
Sep 7 at 8:04
add a comment |Â
Thank you very much!
â Shimmy
Sep 7 at 8:04
Thank you very much!
â Shimmy
Sep 7 at 8:04
Thank you very much!
â Shimmy
Sep 7 at 8:04
add a comment |Â
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