Calculating the variance of a random variable
$begingroup$
I have stumbled upon this question from my text book and I have been finding difficulty in understanding and solving it.
If X and Y are independent random variables with variances $σ_X^2 = 5$ and $σ_Y^2 = 3$. Find the variance of the random variable $Z = −2X + 4Y − 3$. Repeat afterwards this exercise, with $X$ and $Y$ not independent and $σ_{XY} = 1$.
Far as I know, if you have independent variables then $textrm{var}(X+Y)= textrm{var}(X)+textrm{var}(Y)$ also $textrm{var}(XY)=textrm{var}(X)textrm{var}(Y)$.
For the first question I just filled in 5 and 3:
$$Z=-2(5)+4(3)-3=-1$$
For the second question, I have absolutely no idea how to approach it.. Any help would be appreciated
probability random-variables variance
edited Dec 5 '18 at 16:05
gt6989b
34.6k22456
asked Dec 5 '18 at 15:50
Danie BaderDanie Bader
83
$endgroup$
add a comment |
$begingroup$
I have stumbled upon this question from my text book and I have been finding difficulty in understanding and solving it.
If X and Y are independent random variables with variances $σ_X^2 = 5$ and $σ_Y^2 = 3$. Find the variance of the random variable $Z = −2X + 4Y − 3$. Repeat afterwards this exercise, with $X$ and $Y$ not independent and $σ_{XY} = 1$.
Far as I know, if you have independent variables then $textrm{var}(X+Y)= textrm{var}(X)+textrm{var}(Y)$ also $textrm{var}(XY)=textrm{var}(X)textrm{var}(Y)$.
For the first question I just filled in 5 and 3:
$$Z=-2(5)+4(3)-3=-1$$
For the second question, I have absolutely no idea how to approach it.. Any help would be appreciated
probability random-variables variance
edited Dec 5 '18 at 16:05
gt6989b
34.6k22456
asked Dec 5 '18 at 15:50
Danie BaderDanie Bader
83
$endgroup$
1
$begingroup$
Your first answer is wrong. You have to use the fact that $operatorname{var}(aX)=a^2operatorname{var}(X)$.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 15:57
add a comment |
$begingroup$
I have stumbled upon this question from my text book and I have been finding difficulty in understanding and solving it.
If X and Y are independent random variables with variances $σ_X^2 = 5$ and $σ_Y^2 = 3$. Find the variance of the random variable $Z = −2X + 4Y − 3$. Repeat afterwards this exercise, with $X$ and $Y$ not independent and $σ_{XY} = 1$.
Far as I know, if you have independent variables then $textrm{var}(X+Y)= textrm{var}(X)+textrm{var}(Y)$ also $textrm{var}(XY)=textrm{var}(X)textrm{var}(Y)$.
For the first question I just filled in 5 and 3:
$$Z=-2(5)+4(3)-3=-1$$
For the second question, I have absolutely no idea how to approach it.. Any help would be appreciated
probability random-variables variance
edited Dec 5 '18 at 16:05
gt6989b
34.6k22456
asked Dec 5 '18 at 15:50
Danie BaderDanie Bader
83
$endgroup$
I have stumbled upon this question from my text book and I have been finding difficulty in understanding and solving it.
If X and Y are independent random variables with variances $σ_X^2 = 5$ and $σ_Y^2 = 3$. Find the variance of the random variable $Z = −2X + 4Y − 3$. Repeat afterwards this exercise, with $X$ and $Y$ not independent and $σ_{XY} = 1$.
Far as I know, if you have independent variables then $textrm{var}(X+Y)= textrm{var}(X)+textrm{var}(Y)$ also $textrm{var}(XY)=textrm{var}(X)textrm{var}(Y)$.
For the first question I just filled in 5 and 3:
$$Z=-2(5)+4(3)-3=-1$$
For the second question, I have absolutely no idea how to approach it.. Any help would be appreciated
probability random-variables variance
probability random-variables variance
edited Dec 5 '18 at 16:05
gt6989b
34.6k22456
asked Dec 5 '18 at 15:50
Danie BaderDanie Bader
83
edited Dec 5 '18 at 16:05
gt6989b
34.6k22456
asked Dec 5 '18 at 15:50
Danie BaderDanie Bader
83
edited Dec 5 '18 at 16:05
gt6989b
34.6k22456
edited Dec 5 '18 at 16:05
gt6989b
34.6k22456
edited Dec 5 '18 at 16:05
gt6989b
34.6k22456
34.6k22456
asked Dec 5 '18 at 15:50
Danie BaderDanie Bader
83
asked Dec 5 '18 at 15:50
Danie BaderDanie Bader
83
asked Dec 5 '18 at 15:50
Danie BaderDanie Bader
83
83
1
$begingroup$
Your first answer is wrong. You have to use the fact that $operatorname{var}(aX)=a^2operatorname{var}(X)$.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 15:57
add a comment |
1
$begingroup$
Your first answer is wrong. You have to use the fact that $operatorname{var}(aX)=a^2operatorname{var}(X)$.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 15:57
1
1
$begingroup$
Your first answer is wrong. You have to use the fact that $operatorname{var}(aX)=a^2operatorname{var}(X)$.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 15:57
$begingroup$
Your first answer is wrong. You have to use the fact that $operatorname{var}(aX)=a^2operatorname{var}(X)$.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 15:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's do this from the definitions. Note that
$$
Var X = mathbb{E}left[X^2right] - mathbb{E}[X]^2 = m_{X^2} - m_X^2,
$$
with $m$ denoting the expected value for a quick shorthand.
Therefore,
$$
begin{split}
Var (X+Y)
&= mathbb{E}left[(X+Y)^2right] - mathbb{E}[X+Y]^2 \
&= mathbb{E}left[X^2right] + mathbb{E}left[Y^2right] + 2mathbb{E}left[XYright]
- m_X^2 - m_Y^2 - 2m_Xm_Y \
&= sigma_X^2 + sigma_Y^2 + 2(mathbb{E}[XY] - m_X m_Y)
end{split}
$$
Now, if $X,Y$ are independent,
$$
begin{split}
mathbb{E}[XY]
&= iint_mathbb{R^2} xy f_{X,Y}(x,y)dxdy \
&= iint_mathbb{R^2} xy f_X(x) f_Y(y) dxdy \
&= left(int_mathbb{R} xf_X(x) right)left(int_mathbb{R} yf_Y(y) right) \
&= m_X m_Y.
end{split}
$$
This implies $Var(X+Y) = sigma_X^2 + sigma_Y^2$. When $X,Y$ are dependent, that's not necessarily true anymore.
In particular, if you let $X=Y$, it's easy to prove that
$$
Var(X+Y) = Var(X+X) = Var(2X) = 4sigma_X^2.
$$
Can you finish this?
answered Dec 5 '18 at 16:04
gt6989bgt6989b
34.6k22456
$endgroup$
$begingroup$
Thank you! However I still can't understand how I'd make use of the given that Var(XY)=1 to answer the second question
$endgroup$
– Danie Bader
Dec 5 '18 at 16:19
$begingroup$
@DanieBader $sigma_{XY}$ is not variance of $Xcdot Y$, but the covariance, i.e., $$sigma_{XY} = mathbb{E}[(X - m_X)(Y-m_Y)].$$
$endgroup$
– gt6989b
Dec 5 '18 at 16:40
$begingroup$
this makes a lot of sense now.. thank you so much!
$endgroup$
– Danie Bader
Dec 5 '18 at 16:48
add a comment |
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$begingroup$
Let's do this from the definitions. Note that
$$
Var X = mathbb{E}left[X^2right] - mathbb{E}[X]^2 = m_{X^2} - m_X^2,
$$
with $m$ denoting the expected value for a quick shorthand.
Therefore,
$$
begin{split}
Var (X+Y)
&= mathbb{E}left[(X+Y)^2right] - mathbb{E}[X+Y]^2 \
&= mathbb{E}left[X^2right] + mathbb{E}left[Y^2right] + 2mathbb{E}left[XYright]
- m_X^2 - m_Y^2 - 2m_Xm_Y \
&= sigma_X^2 + sigma_Y^2 + 2(mathbb{E}[XY] - m_X m_Y)
end{split}
$$
Now, if $X,Y$ are independent,
$$
begin{split}
mathbb{E}[XY]
&= iint_mathbb{R^2} xy f_{X,Y}(x,y)dxdy \
&= iint_mathbb{R^2} xy f_X(x) f_Y(y) dxdy \
&= left(int_mathbb{R} xf_X(x) right)left(int_mathbb{R} yf_Y(y) right) \
&= m_X m_Y.
end{split}
$$
This implies $Var(X+Y) = sigma_X^2 + sigma_Y^2$. When $X,Y$ are dependent, that's not necessarily true anymore.
In particular, if you let $X=Y$, it's easy to prove that
$$
Var(X+Y) = Var(X+X) = Var(2X) = 4sigma_X^2.
$$
Can you finish this?
answered Dec 5 '18 at 16:04
gt6989bgt6989b
34.6k22456
$endgroup$
$begingroup$
Thank you! However I still can't understand how I'd make use of the given that Var(XY)=1 to answer the second question
$endgroup$
– Danie Bader
Dec 5 '18 at 16:19
$begingroup$
@DanieBader $sigma_{XY}$ is not variance of $Xcdot Y$, but the covariance, i.e., $$sigma_{XY} = mathbb{E}[(X - m_X)(Y-m_Y)].$$
$endgroup$
– gt6989b
Dec 5 '18 at 16:40
$begingroup$
this makes a lot of sense now.. thank you so much!
$endgroup$
– Danie Bader
Dec 5 '18 at 16:48
add a comment |
$begingroup$
Let's do this from the definitions. Note that
$$
Var X = mathbb{E}left[X^2right] - mathbb{E}[X]^2 = m_{X^2} - m_X^2,
$$
with $m$ denoting the expected value for a quick shorthand.
Therefore,
$$
begin{split}
Var (X+Y)
&= mathbb{E}left[(X+Y)^2right] - mathbb{E}[X+Y]^2 \
&= mathbb{E}left[X^2right] + mathbb{E}left[Y^2right] + 2mathbb{E}left[XYright]
- m_X^2 - m_Y^2 - 2m_Xm_Y \
&= sigma_X^2 + sigma_Y^2 + 2(mathbb{E}[XY] - m_X m_Y)
end{split}
$$
Now, if $X,Y$ are independent,
$$
begin{split}
mathbb{E}[XY]
&= iint_mathbb{R^2} xy f_{X,Y}(x,y)dxdy \
&= iint_mathbb{R^2} xy f_X(x) f_Y(y) dxdy \
&= left(int_mathbb{R} xf_X(x) right)left(int_mathbb{R} yf_Y(y) right) \
&= m_X m_Y.
end{split}
$$
This implies $Var(X+Y) = sigma_X^2 + sigma_Y^2$. When $X,Y$ are dependent, that's not necessarily true anymore.
In particular, if you let $X=Y$, it's easy to prove that
$$
Var(X+Y) = Var(X+X) = Var(2X) = 4sigma_X^2.
$$
Can you finish this?
answered Dec 5 '18 at 16:04
gt6989bgt6989b
34.6k22456
$endgroup$
$begingroup$
Thank you! However I still can't understand how I'd make use of the given that Var(XY)=1 to answer the second question
$endgroup$
– Danie Bader
Dec 5 '18 at 16:19
$begingroup$
@DanieBader $sigma_{XY}$ is not variance of $Xcdot Y$, but the covariance, i.e., $$sigma_{XY} = mathbb{E}[(X - m_X)(Y-m_Y)].$$
$endgroup$
– gt6989b
Dec 5 '18 at 16:40
$begingroup$
this makes a lot of sense now.. thank you so much!
$endgroup$
– Danie Bader
Dec 5 '18 at 16:48
add a comment |
$begingroup$
Let's do this from the definitions. Note that
$$
Var X = mathbb{E}left[X^2right] - mathbb{E}[X]^2 = m_{X^2} - m_X^2,
$$
with $m$ denoting the expected value for a quick shorthand.
Therefore,
$$
begin{split}
Var (X+Y)
&= mathbb{E}left[(X+Y)^2right] - mathbb{E}[X+Y]^2 \
&= mathbb{E}left[X^2right] + mathbb{E}left[Y^2right] + 2mathbb{E}left[XYright]
- m_X^2 - m_Y^2 - 2m_Xm_Y \
&= sigma_X^2 + sigma_Y^2 + 2(mathbb{E}[XY] - m_X m_Y)
end{split}
$$
Now, if $X,Y$ are independent,
$$
begin{split}
mathbb{E}[XY]
&= iint_mathbb{R^2} xy f_{X,Y}(x,y)dxdy \
&= iint_mathbb{R^2} xy f_X(x) f_Y(y) dxdy \
&= left(int_mathbb{R} xf_X(x) right)left(int_mathbb{R} yf_Y(y) right) \
&= m_X m_Y.
end{split}
$$
This implies $Var(X+Y) = sigma_X^2 + sigma_Y^2$. When $X,Y$ are dependent, that's not necessarily true anymore.
In particular, if you let $X=Y$, it's easy to prove that
$$
Var(X+Y) = Var(X+X) = Var(2X) = 4sigma_X^2.
$$
Can you finish this?
answered Dec 5 '18 at 16:04
gt6989bgt6989b
34.6k22456
$endgroup$
Let's do this from the definitions. Note that
$$
Var X = mathbb{E}left[X^2right] - mathbb{E}[X]^2 = m_{X^2} - m_X^2,
$$
with $m$ denoting the expected value for a quick shorthand.
Therefore,
$$
begin{split}
Var (X+Y)
&= mathbb{E}left[(X+Y)^2right] - mathbb{E}[X+Y]^2 \
&= mathbb{E}left[X^2right] + mathbb{E}left[Y^2right] + 2mathbb{E}left[XYright]
- m_X^2 - m_Y^2 - 2m_Xm_Y \
&= sigma_X^2 + sigma_Y^2 + 2(mathbb{E}[XY] - m_X m_Y)
end{split}
$$
Now, if $X,Y$ are independent,
$$
begin{split}
mathbb{E}[XY]
&= iint_mathbb{R^2} xy f_{X,Y}(x,y)dxdy \
&= iint_mathbb{R^2} xy f_X(x) f_Y(y) dxdy \
&= left(int_mathbb{R} xf_X(x) right)left(int_mathbb{R} yf_Y(y) right) \
&= m_X m_Y.
end{split}
$$
This implies $Var(X+Y) = sigma_X^2 + sigma_Y^2$. When $X,Y$ are dependent, that's not necessarily true anymore.
In particular, if you let $X=Y$, it's easy to prove that
$$
Var(X+Y) = Var(X+X) = Var(2X) = 4sigma_X^2.
$$
Can you finish this?
answered Dec 5 '18 at 16:04
gt6989bgt6989b
34.6k22456
answered Dec 5 '18 at 16:04
gt6989bgt6989b
34.6k22456
answered Dec 5 '18 at 16:04
gt6989bgt6989b
34.6k22456
answered Dec 5 '18 at 16:04
gt6989bgt6989b
34.6k22456
34.6k22456
$begingroup$
Thank you! However I still can't understand how I'd make use of the given that Var(XY)=1 to answer the second question
$endgroup$
– Danie Bader
Dec 5 '18 at 16:19
$begingroup$
@DanieBader $sigma_{XY}$ is not variance of $Xcdot Y$, but the covariance, i.e., $$sigma_{XY} = mathbb{E}[(X - m_X)(Y-m_Y)].$$
$endgroup$
– gt6989b
Dec 5 '18 at 16:40
$begingroup$
this makes a lot of sense now.. thank you so much!
$endgroup$
– Danie Bader
Dec 5 '18 at 16:48
add a comment |
$begingroup$
Thank you! However I still can't understand how I'd make use of the given that Var(XY)=1 to answer the second question
$endgroup$
– Danie Bader
Dec 5 '18 at 16:19
$begingroup$
@DanieBader $sigma_{XY}$ is not variance of $Xcdot Y$, but the covariance, i.e., $$sigma_{XY} = mathbb{E}[(X - m_X)(Y-m_Y)].$$
$endgroup$
– gt6989b
Dec 5 '18 at 16:40
$begingroup$
this makes a lot of sense now.. thank you so much!
$endgroup$
– Danie Bader
Dec 5 '18 at 16:48
$begingroup$
Thank you! However I still can't understand how I'd make use of the given that Var(XY)=1 to answer the second question
$endgroup$
– Danie Bader
Dec 5 '18 at 16:19
$begingroup$
Thank you! However I still can't understand how I'd make use of the given that Var(XY)=1 to answer the second question
$endgroup$
– Danie Bader
Dec 5 '18 at 16:19
$begingroup$
@DanieBader $sigma_{XY}$ is not variance of $Xcdot Y$, but the covariance, i.e., $$sigma_{XY} = mathbb{E}[(X - m_X)(Y-m_Y)].$$
$endgroup$
– gt6989b
Dec 5 '18 at 16:40
$begingroup$
@DanieBader $sigma_{XY}$ is not variance of $Xcdot Y$, but the covariance, i.e., $$sigma_{XY} = mathbb{E}[(X - m_X)(Y-m_Y)].$$
$endgroup$
– gt6989b
Dec 5 '18 at 16:40
$begingroup$
this makes a lot of sense now.. thank you so much!
$endgroup$
– Danie Bader
Dec 5 '18 at 16:48
$begingroup$
this makes a lot of sense now.. thank you so much!
$endgroup$
– Danie Bader
Dec 5 '18 at 16:48
add a comment |
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$begingroup$
Your first answer is wrong. You have to use the fact that $operatorname{var}(aX)=a^2operatorname{var}(X)$.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 15:57