Mixing
Adding Independent Random Variables Given Their Individual Expectations and Variance
$begingroup$
How do I add or subtract independent random variables (R.Vs) when given their individual expectations and variance?
I'm a student in high school and I haven't covered distributions yet, so please try not to use them.
Example, R.Vs A, B & C
Where
$E(A)= 35;; Var(A)=8\
E(B) = 25;;;; Var(B)=9\$
Calculate the expectation and variance of:
$A + 2B$
probability random-variables variance expected-value
asked Jan 28 at 1:35
landlockedorcalandlockedorca
83
$endgroup$
add a comment |
$begingroup$
How do I add or subtract independent random variables (R.Vs) when given their individual expectations and variance?
I'm a student in high school and I haven't covered distributions yet, so please try not to use them.
Example, R.Vs A, B & C
Where
$E(A)= 35;; Var(A)=8\
E(B) = 25;;;; Var(B)=9\$
Calculate the expectation and variance of:
$A + 2B$
probability random-variables variance expected-value
asked Jan 28 at 1:35
landlockedorcalandlockedorca
83
$endgroup$
add a comment |
$begingroup$
How do I add or subtract independent random variables (R.Vs) when given their individual expectations and variance?
I'm a student in high school and I haven't covered distributions yet, so please try not to use them.
Example, R.Vs A, B & C
Where
$E(A)= 35;; Var(A)=8\
E(B) = 25;;;; Var(B)=9\$
Calculate the expectation and variance of:
$A + 2B$
probability random-variables variance expected-value
asked Jan 28 at 1:35
landlockedorcalandlockedorca
83
$endgroup$
How do I add or subtract independent random variables (R.Vs) when given their individual expectations and variance?
I'm a student in high school and I haven't covered distributions yet, so please try not to use them.
Example, R.Vs A, B & C
Where
$E(A)= 35;; Var(A)=8\
E(B) = 25;;;; Var(B)=9\$
Calculate the expectation and variance of:
$A + 2B$
probability random-variables variance expected-value
probability random-variables variance expected-value
asked Jan 28 at 1:35
landlockedorcalandlockedorca
83
asked Jan 28 at 1:35
landlockedorcalandlockedorca
83
edited Jan 28 at 10:00
landlockedorca
asked Jan 28 at 1:35
landlockedorcalandlockedorca
83
asked Jan 28 at 1:35
landlockedorcalandlockedorca
83
asked Jan 28 at 1:35
landlockedorcalandlockedorca
83
83
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $A$ and $B$ be two random variables and $c$ be a constant. Then,
$mathbb{E}[A + cB] = mathbb{E}[A] + cmathbb{E}[B]$ and
$operatorname{Var}(A + cB) = operatorname{Var}(A) + c^2 operatorname{Var}(B)$ (assuming $A$ and $B$ are independent).
Variance is defined in terms of the expectation. In particular, $operatorname{Var}(X) = mathbb{E}[(X - mathbb{E}[X])^2]$. See if you can use this definition to prove property (2) from property (1).
answered Jan 28 at 1:48
parsiadparsiad
18.6k32453
$endgroup$
add a comment |
$begingroup$
Expectation is a linear function. So,
$$E(sum_{i=1}^n k_iX_i)=sum_{i=1}^nk_iE(X_i)$$
Variance is " just like " squaring and covariance is "just like" multiplication. So, we can easily expand variance using identities of squares. In particular,
$$V(X_1+X_2)=V(X_1)+V(X_2)+2Cov(X_1,X_2)$$
which is analogous to $(a+b)^2=a^2+b^2+2ab$.
Similarly,
$$V(sum_{i=1}^n k_iX_i)=sum_{i=1}^nk_i^2V(X_i) + 2sum_{i=1}^nsum_{j=1}^{i-1}k_ik_jCov(X_iX_j)$$
Hope it is helpful
answered Jan 28 at 2:03
MartundMartund
1,667213
$endgroup$
$begingroup$
The section on variance using identities of squares is very clear. I haven't done the notation which you're using at the start and at the end yet, so I have no clue what you mean, sorry.
$endgroup$
– landlockedorca
Jan 28 at 9:35
$begingroup$
Which notation you are not getting?
$endgroup$
– Martund
Jan 29 at 2:39
$begingroup$
I've neither done covariance, nor have I seen the n and the i = 1 above the sum sign before.
$endgroup$
– landlockedorca
Jan 30 at 8:34
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
Let $A$ and $B$ be two random variables and $c$ be a constant. Then,
$mathbb{E}[A + cB] = mathbb{E}[A] + cmathbb{E}[B]$ and
$operatorname{Var}(A + cB) = operatorname{Var}(A) + c^2 operatorname{Var}(B)$ (assuming $A$ and $B$ are independent).
Variance is defined in terms of the expectation. In particular, $operatorname{Var}(X) = mathbb{E}[(X - mathbb{E}[X])^2]$. See if you can use this definition to prove property (2) from property (1).
answered Jan 28 at 1:48
parsiadparsiad
18.6k32453
$endgroup$
add a comment |
$begingroup$
Let $A$ and $B$ be two random variables and $c$ be a constant. Then,
$mathbb{E}[A + cB] = mathbb{E}[A] + cmathbb{E}[B]$ and
$operatorname{Var}(A + cB) = operatorname{Var}(A) + c^2 operatorname{Var}(B)$ (assuming $A$ and $B$ are independent).
Variance is defined in terms of the expectation. In particular, $operatorname{Var}(X) = mathbb{E}[(X - mathbb{E}[X])^2]$. See if you can use this definition to prove property (2) from property (1).
answered Jan 28 at 1:48
parsiadparsiad
18.6k32453
$endgroup$
add a comment |
$begingroup$
Let $A$ and $B$ be two random variables and $c$ be a constant. Then,
$mathbb{E}[A + cB] = mathbb{E}[A] + cmathbb{E}[B]$ and
$operatorname{Var}(A + cB) = operatorname{Var}(A) + c^2 operatorname{Var}(B)$ (assuming $A$ and $B$ are independent).
Variance is defined in terms of the expectation. In particular, $operatorname{Var}(X) = mathbb{E}[(X - mathbb{E}[X])^2]$. See if you can use this definition to prove property (2) from property (1).
answered Jan 28 at 1:48
parsiadparsiad
18.6k32453
$endgroup$
Let $A$ and $B$ be two random variables and $c$ be a constant. Then,
$mathbb{E}[A + cB] = mathbb{E}[A] + cmathbb{E}[B]$ and
$operatorname{Var}(A + cB) = operatorname{Var}(A) + c^2 operatorname{Var}(B)$ (assuming $A$ and $B$ are independent).
Variance is defined in terms of the expectation. In particular, $operatorname{Var}(X) = mathbb{E}[(X - mathbb{E}[X])^2]$. See if you can use this definition to prove property (2) from property (1).
answered Jan 28 at 1:48
parsiadparsiad
18.6k32453
edited Jan 28 at 5:46
answered Jan 28 at 1:48
parsiadparsiad
18.6k32453
answered Jan 28 at 1:48
parsiadparsiad
18.6k32453
answered Jan 28 at 1:48
parsiadparsiad
18.6k32453
18.6k32453
add a comment |
add a comment |
$begingroup$
Expectation is a linear function. So,
$$E(sum_{i=1}^n k_iX_i)=sum_{i=1}^nk_iE(X_i)$$
Variance is " just like " squaring and covariance is "just like" multiplication. So, we can easily expand variance using identities of squares. In particular,
$$V(X_1+X_2)=V(X_1)+V(X_2)+2Cov(X_1,X_2)$$
which is analogous to $(a+b)^2=a^2+b^2+2ab$.
Similarly,
$$V(sum_{i=1}^n k_iX_i)=sum_{i=1}^nk_i^2V(X_i) + 2sum_{i=1}^nsum_{j=1}^{i-1}k_ik_jCov(X_iX_j)$$
Hope it is helpful
answered Jan 28 at 2:03
MartundMartund
1,667213
$endgroup$
$begingroup$
The section on variance using identities of squares is very clear. I haven't done the notation which you're using at the start and at the end yet, so I have no clue what you mean, sorry.
$endgroup$
– landlockedorca
Jan 28 at 9:35
$begingroup$
Which notation you are not getting?
$endgroup$
– Martund
Jan 29 at 2:39
$begingroup$
I've neither done covariance, nor have I seen the n and the i = 1 above the sum sign before.
$endgroup$
– landlockedorca
Jan 30 at 8:34
add a comment |
$begingroup$
Expectation is a linear function. So,
$$E(sum_{i=1}^n k_iX_i)=sum_{i=1}^nk_iE(X_i)$$
Variance is " just like " squaring and covariance is "just like" multiplication. So, we can easily expand variance using identities of squares. In particular,
$$V(X_1+X_2)=V(X_1)+V(X_2)+2Cov(X_1,X_2)$$
which is analogous to $(a+b)^2=a^2+b^2+2ab$.
Similarly,
$$V(sum_{i=1}^n k_iX_i)=sum_{i=1}^nk_i^2V(X_i) + 2sum_{i=1}^nsum_{j=1}^{i-1}k_ik_jCov(X_iX_j)$$
Hope it is helpful
answered Jan 28 at 2:03
MartundMartund
1,667213
$endgroup$
$begingroup$
The section on variance using identities of squares is very clear. I haven't done the notation which you're using at the start and at the end yet, so I have no clue what you mean, sorry.
$endgroup$
– landlockedorca
Jan 28 at 9:35
$begingroup$
Which notation you are not getting?
$endgroup$
– Martund
Jan 29 at 2:39
$begingroup$
I've neither done covariance, nor have I seen the n and the i = 1 above the sum sign before.
$endgroup$
– landlockedorca
Jan 30 at 8:34
add a comment |
$begingroup$
Expectation is a linear function. So,
$$E(sum_{i=1}^n k_iX_i)=sum_{i=1}^nk_iE(X_i)$$
Variance is " just like " squaring and covariance is "just like" multiplication. So, we can easily expand variance using identities of squares. In particular,
$$V(X_1+X_2)=V(X_1)+V(X_2)+2Cov(X_1,X_2)$$
which is analogous to $(a+b)^2=a^2+b^2+2ab$.
Similarly,
$$V(sum_{i=1}^n k_iX_i)=sum_{i=1}^nk_i^2V(X_i) + 2sum_{i=1}^nsum_{j=1}^{i-1}k_ik_jCov(X_iX_j)$$
Hope it is helpful
answered Jan 28 at 2:03
MartundMartund
1,667213
$endgroup$
Expectation is a linear function. So,
$$E(sum_{i=1}^n k_iX_i)=sum_{i=1}^nk_iE(X_i)$$
Variance is " just like " squaring and covariance is "just like" multiplication. So, we can easily expand variance using identities of squares. In particular,
$$V(X_1+X_2)=V(X_1)+V(X_2)+2Cov(X_1,X_2)$$
which is analogous to $(a+b)^2=a^2+b^2+2ab$.
Similarly,
$$V(sum_{i=1}^n k_iX_i)=sum_{i=1}^nk_i^2V(X_i) + 2sum_{i=1}^nsum_{j=1}^{i-1}k_ik_jCov(X_iX_j)$$
Hope it is helpful
answered Jan 28 at 2:03
MartundMartund
1,667213
answered Jan 28 at 2:03
MartundMartund
1,667213
answered Jan 28 at 2:03
MartundMartund
1,667213
answered Jan 28 at 2:03
MartundMartund
1,667213
1,667213
$begingroup$
The section on variance using identities of squares is very clear. I haven't done the notation which you're using at the start and at the end yet, so I have no clue what you mean, sorry.
$endgroup$
– landlockedorca
Jan 28 at 9:35
$begingroup$
Which notation you are not getting?
$endgroup$
– Martund
Jan 29 at 2:39
$begingroup$
I've neither done covariance, nor have I seen the n and the i = 1 above the sum sign before.
$endgroup$
– landlockedorca
Jan 30 at 8:34
add a comment |
$begingroup$
The section on variance using identities of squares is very clear. I haven't done the notation which you're using at the start and at the end yet, so I have no clue what you mean, sorry.
$endgroup$
– landlockedorca
Jan 28 at 9:35
$begingroup$
Which notation you are not getting?
$endgroup$
– Martund
Jan 29 at 2:39
$begingroup$
I've neither done covariance, nor have I seen the n and the i = 1 above the sum sign before.
$endgroup$
– landlockedorca
Jan 30 at 8:34
$begingroup$
The section on variance using identities of squares is very clear. I haven't done the notation which you're using at the start and at the end yet, so I have no clue what you mean, sorry.
$endgroup$
– landlockedorca
Jan 28 at 9:35
$begingroup$
The section on variance using identities of squares is very clear. I haven't done the notation which you're using at the start and at the end yet, so I have no clue what you mean, sorry.
$endgroup$
– landlockedorca
Jan 28 at 9:35
$begingroup$
Which notation you are not getting?
$endgroup$
– Martund
Jan 29 at 2:39
$begingroup$
Which notation you are not getting?
$endgroup$
– Martund
Jan 29 at 2:39
$begingroup$
I've neither done covariance, nor have I seen the n and the i = 1 above the sum sign before.
$endgroup$
– landlockedorca
Jan 30 at 8:34
$begingroup$
I've neither done covariance, nor have I seen the n and the i = 1 above the sum sign before.
$endgroup$
– landlockedorca
Jan 30 at 8:34
add a comment |
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