Mixing
Variance of a mixed variable
$begingroup$
I am studying probability on my own and couldn't find anything on this in my textbooks.
Say we have two random variables, $X$ and $Y$, with known means $mu_{1}$ and $mu_{2}$, and known variances $sigma^{2}_{1}$ and $sigma^{2}_{2}$.
$A$ is defined to be a linear combination of both variables. How can you interpret the variance of $A$? And how can you calculate $textbf{E}(A^{2})$ in order to find such variance?
Does this generalize to more complex functions $A(X,Y)$?
probability variance
edited Feb 3 at 1:12
APC89
2,371720
asked Feb 3 at 0:59
Kumail AlhamoudKumail Alhamoud
83
$endgroup$
add a comment |
$begingroup$
I am studying probability on my own and couldn't find anything on this in my textbooks.
Say we have two random variables, $X$ and $Y$, with known means $mu_{1}$ and $mu_{2}$, and known variances $sigma^{2}_{1}$ and $sigma^{2}_{2}$.
$A$ is defined to be a linear combination of both variables. How can you interpret the variance of $A$? And how can you calculate $textbf{E}(A^{2})$ in order to find such variance?
Does this generalize to more complex functions $A(X,Y)$?
probability variance
edited Feb 3 at 1:12
APC89
2,371720
asked Feb 3 at 0:59
Kumail AlhamoudKumail Alhamoud
83
$endgroup$
$begingroup$
Concerning the calculation of the variance of a sum of random variables, search for "Bienaymé formula". Concerning the interpretation: What exactly are you thinking of? Do you have a general understanding of what the variance of a random variable is? Why should it be different for a sum? Concerning generelizations: Any formula you know for $E(f(X))$ (e.g. if you know a density) can be applied to calculate the variance. Is this what you have in mind?
$endgroup$
– Mars Plastic
Feb 3 at 1:21
$begingroup$
But in general, you cannot expect to calculate the variance of $X+Y$ just from the variances of $X$ and $Y$. If for example $X=Y$, we have $Var(X+Y)=4Var(X)$, but if they have the same distribution but are independent, we have $Var(X+Y)=2Var(X)$
$endgroup$
– Mars Plastic
Feb 3 at 1:25
$begingroup$
I guess I was not paying attention to the role of independence in how the sum might vary. Taking independence into account, I think I now see how the meaning of a variance of a sum not that different. Can you direct me to a link to learn more about expectations of functions of random variables?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:56
add a comment |
$begingroup$
I am studying probability on my own and couldn't find anything on this in my textbooks.
Say we have two random variables, $X$ and $Y$, with known means $mu_{1}$ and $mu_{2}$, and known variances $sigma^{2}_{1}$ and $sigma^{2}_{2}$.
$A$ is defined to be a linear combination of both variables. How can you interpret the variance of $A$? And how can you calculate $textbf{E}(A^{2})$ in order to find such variance?
Does this generalize to more complex functions $A(X,Y)$?
probability variance
edited Feb 3 at 1:12
APC89
2,371720
asked Feb 3 at 0:59
Kumail AlhamoudKumail Alhamoud
83
$endgroup$
I am studying probability on my own and couldn't find anything on this in my textbooks.
Say we have two random variables, $X$ and $Y$, with known means $mu_{1}$ and $mu_{2}$, and known variances $sigma^{2}_{1}$ and $sigma^{2}_{2}$.
$A$ is defined to be a linear combination of both variables. How can you interpret the variance of $A$? And how can you calculate $textbf{E}(A^{2})$ in order to find such variance?
Does this generalize to more complex functions $A(X,Y)$?
probability variance
probability variance
edited Feb 3 at 1:12
APC89
2,371720
asked Feb 3 at 0:59
Kumail AlhamoudKumail Alhamoud
83
edited Feb 3 at 1:12
APC89
2,371720
asked Feb 3 at 0:59
Kumail AlhamoudKumail Alhamoud
83
edited Feb 3 at 1:12
APC89
2,371720
edited Feb 3 at 1:12
APC89
2,371720
edited Feb 3 at 1:12
APC89
2,371720
2,371720
asked Feb 3 at 0:59
Kumail AlhamoudKumail Alhamoud
83
asked Feb 3 at 0:59
Kumail AlhamoudKumail Alhamoud
83
asked Feb 3 at 0:59
Kumail AlhamoudKumail Alhamoud
83
83
$begingroup$
Concerning the calculation of the variance of a sum of random variables, search for "Bienaymé formula". Concerning the interpretation: What exactly are you thinking of? Do you have a general understanding of what the variance of a random variable is? Why should it be different for a sum? Concerning generelizations: Any formula you know for $E(f(X))$ (e.g. if you know a density) can be applied to calculate the variance. Is this what you have in mind?
$endgroup$
– Mars Plastic
Feb 3 at 1:21
$begingroup$
But in general, you cannot expect to calculate the variance of $X+Y$ just from the variances of $X$ and $Y$. If for example $X=Y$, we have $Var(X+Y)=4Var(X)$, but if they have the same distribution but are independent, we have $Var(X+Y)=2Var(X)$
$endgroup$
– Mars Plastic
Feb 3 at 1:25
$begingroup$
I guess I was not paying attention to the role of independence in how the sum might vary. Taking independence into account, I think I now see how the meaning of a variance of a sum not that different. Can you direct me to a link to learn more about expectations of functions of random variables?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:56
add a comment |
$begingroup$
Concerning the calculation of the variance of a sum of random variables, search for "Bienaymé formula". Concerning the interpretation: What exactly are you thinking of? Do you have a general understanding of what the variance of a random variable is? Why should it be different for a sum? Concerning generelizations: Any formula you know for $E(f(X))$ (e.g. if you know a density) can be applied to calculate the variance. Is this what you have in mind?
$endgroup$
– Mars Plastic
Feb 3 at 1:21
$begingroup$
But in general, you cannot expect to calculate the variance of $X+Y$ just from the variances of $X$ and $Y$. If for example $X=Y$, we have $Var(X+Y)=4Var(X)$, but if they have the same distribution but are independent, we have $Var(X+Y)=2Var(X)$
$endgroup$
– Mars Plastic
Feb 3 at 1:25
$begingroup$
I guess I was not paying attention to the role of independence in how the sum might vary. Taking independence into account, I think I now see how the meaning of a variance of a sum not that different. Can you direct me to a link to learn more about expectations of functions of random variables?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:56
$begingroup$
Concerning the calculation of the variance of a sum of random variables, search for "Bienaymé formula". Concerning the interpretation: What exactly are you thinking of? Do you have a general understanding of what the variance of a random variable is? Why should it be different for a sum? Concerning generelizations: Any formula you know for $E(f(X))$ (e.g. if you know a density) can be applied to calculate the variance. Is this what you have in mind?
$endgroup$
– Mars Plastic
Feb 3 at 1:21
$begingroup$
Concerning the calculation of the variance of a sum of random variables, search for "Bienaymé formula". Concerning the interpretation: What exactly are you thinking of? Do you have a general understanding of what the variance of a random variable is? Why should it be different for a sum? Concerning generelizations: Any formula you know for $E(f(X))$ (e.g. if you know a density) can be applied to calculate the variance. Is this what you have in mind?
$endgroup$
– Mars Plastic
Feb 3 at 1:21
$begingroup$
But in general, you cannot expect to calculate the variance of $X+Y$ just from the variances of $X$ and $Y$. If for example $X=Y$, we have $Var(X+Y)=4Var(X)$, but if they have the same distribution but are independent, we have $Var(X+Y)=2Var(X)$
$endgroup$
– Mars Plastic
Feb 3 at 1:25
$begingroup$
But in general, you cannot expect to calculate the variance of $X+Y$ just from the variances of $X$ and $Y$. If for example $X=Y$, we have $Var(X+Y)=4Var(X)$, but if they have the same distribution but are independent, we have $Var(X+Y)=2Var(X)$
$endgroup$
– Mars Plastic
Feb 3 at 1:25
$begingroup$
I guess I was not paying attention to the role of independence in how the sum might vary. Taking independence into account, I think I now see how the meaning of a variance of a sum not that different. Can you direct me to a link to learn more about expectations of functions of random variables?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:56
$begingroup$
I guess I was not paying attention to the role of independence in how the sum might vary. Taking independence into account, I think I now see how the meaning of a variance of a sum not that different. Can you direct me to a link to learn more about expectations of functions of random variables?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
According to the data provided and the properties of variance, we have
begin{align*}
textbf{Var}(A) = textbf{Var}(aX+bY) = a^{2}textbf{Var}(X) + b^{2}textbf{Var}(Y) = a^{2}sigma^{2}_{1} + b^{2}sigma^{2}_{2}
end{align*}
Once $X$ and $Y$ are independent, they are uncorrelated, which means that $textbf{Cov}(X,Y) = 0$ and, consequently, the formula above is correct.
As you may have noticed, we did not need to use the value $textbf{E}(A^{2})$ directly in order to obtain the variance: it is enough to make use of the variance properties.
As to the general case, the strategy to solve the problem depends on the expression of $A(X,Y)$.
answered Feb 3 at 1:26
APC89APC89
2,371720
$endgroup$
$begingroup$
I see how the general case might vary. As for 𝐄(𝐴^2), in general, would you just expand the square of A and then see if you get something you can work with?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:59
$begingroup$
In accordance to the given information, a good start point would be to notice that $$textbf{Var}(A) = textbf{E}(A^{2}) - textbf{E}(A)^{2}$$ Once we know the value of $textbf{Var}(A)$ and $textbf{E}(A)$ can be easily calculated, you obtain the value of $textbf{E}(A^{2})$. More precisely, we have $$textbf{E}(A) = textbf{E}(aX+bY) = atextbf{E}(X) + btextbf{E}(Y) = amu_{1} + bmu_{2}$$
$endgroup$
– APC89
Feb 3 at 2:05
add a comment |
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1 Answer
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$begingroup$
According to the data provided and the properties of variance, we have
begin{align*}
textbf{Var}(A) = textbf{Var}(aX+bY) = a^{2}textbf{Var}(X) + b^{2}textbf{Var}(Y) = a^{2}sigma^{2}_{1} + b^{2}sigma^{2}_{2}
end{align*}
Once $X$ and $Y$ are independent, they are uncorrelated, which means that $textbf{Cov}(X,Y) = 0$ and, consequently, the formula above is correct.
As you may have noticed, we did not need to use the value $textbf{E}(A^{2})$ directly in order to obtain the variance: it is enough to make use of the variance properties.
As to the general case, the strategy to solve the problem depends on the expression of $A(X,Y)$.
answered Feb 3 at 1:26
APC89APC89
2,371720
$endgroup$
$begingroup$
I see how the general case might vary. As for 𝐄(𝐴^2), in general, would you just expand the square of A and then see if you get something you can work with?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:59
$begingroup$
In accordance to the given information, a good start point would be to notice that $$textbf{Var}(A) = textbf{E}(A^{2}) - textbf{E}(A)^{2}$$ Once we know the value of $textbf{Var}(A)$ and $textbf{E}(A)$ can be easily calculated, you obtain the value of $textbf{E}(A^{2})$. More precisely, we have $$textbf{E}(A) = textbf{E}(aX+bY) = atextbf{E}(X) + btextbf{E}(Y) = amu_{1} + bmu_{2}$$
$endgroup$
– APC89
Feb 3 at 2:05
add a comment |
$begingroup$
According to the data provided and the properties of variance, we have
begin{align*}
textbf{Var}(A) = textbf{Var}(aX+bY) = a^{2}textbf{Var}(X) + b^{2}textbf{Var}(Y) = a^{2}sigma^{2}_{1} + b^{2}sigma^{2}_{2}
end{align*}
Once $X$ and $Y$ are independent, they are uncorrelated, which means that $textbf{Cov}(X,Y) = 0$ and, consequently, the formula above is correct.
As you may have noticed, we did not need to use the value $textbf{E}(A^{2})$ directly in order to obtain the variance: it is enough to make use of the variance properties.
As to the general case, the strategy to solve the problem depends on the expression of $A(X,Y)$.
answered Feb 3 at 1:26
APC89APC89
2,371720
$endgroup$
$begingroup$
I see how the general case might vary. As for 𝐄(𝐴^2), in general, would you just expand the square of A and then see if you get something you can work with?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:59
$begingroup$
In accordance to the given information, a good start point would be to notice that $$textbf{Var}(A) = textbf{E}(A^{2}) - textbf{E}(A)^{2}$$ Once we know the value of $textbf{Var}(A)$ and $textbf{E}(A)$ can be easily calculated, you obtain the value of $textbf{E}(A^{2})$. More precisely, we have $$textbf{E}(A) = textbf{E}(aX+bY) = atextbf{E}(X) + btextbf{E}(Y) = amu_{1} + bmu_{2}$$
$endgroup$
– APC89
Feb 3 at 2:05
add a comment |
$begingroup$
According to the data provided and the properties of variance, we have
begin{align*}
textbf{Var}(A) = textbf{Var}(aX+bY) = a^{2}textbf{Var}(X) + b^{2}textbf{Var}(Y) = a^{2}sigma^{2}_{1} + b^{2}sigma^{2}_{2}
end{align*}
Once $X$ and $Y$ are independent, they are uncorrelated, which means that $textbf{Cov}(X,Y) = 0$ and, consequently, the formula above is correct.
As you may have noticed, we did not need to use the value $textbf{E}(A^{2})$ directly in order to obtain the variance: it is enough to make use of the variance properties.
As to the general case, the strategy to solve the problem depends on the expression of $A(X,Y)$.
answered Feb 3 at 1:26
APC89APC89
2,371720
$endgroup$
According to the data provided and the properties of variance, we have
begin{align*}
textbf{Var}(A) = textbf{Var}(aX+bY) = a^{2}textbf{Var}(X) + b^{2}textbf{Var}(Y) = a^{2}sigma^{2}_{1} + b^{2}sigma^{2}_{2}
end{align*}
Once $X$ and $Y$ are independent, they are uncorrelated, which means that $textbf{Cov}(X,Y) = 0$ and, consequently, the formula above is correct.
As you may have noticed, we did not need to use the value $textbf{E}(A^{2})$ directly in order to obtain the variance: it is enough to make use of the variance properties.
As to the general case, the strategy to solve the problem depends on the expression of $A(X,Y)$.
answered Feb 3 at 1:26
APC89APC89
2,371720
answered Feb 3 at 1:26
APC89APC89
2,371720
answered Feb 3 at 1:26
APC89APC89
2,371720
answered Feb 3 at 1:26
APC89APC89
2,371720
2,371720
$begingroup$
I see how the general case might vary. As for 𝐄(𝐴^2), in general, would you just expand the square of A and then see if you get something you can work with?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:59
$begingroup$
In accordance to the given information, a good start point would be to notice that $$textbf{Var}(A) = textbf{E}(A^{2}) - textbf{E}(A)^{2}$$ Once we know the value of $textbf{Var}(A)$ and $textbf{E}(A)$ can be easily calculated, you obtain the value of $textbf{E}(A^{2})$. More precisely, we have $$textbf{E}(A) = textbf{E}(aX+bY) = atextbf{E}(X) + btextbf{E}(Y) = amu_{1} + bmu_{2}$$
$endgroup$
– APC89
Feb 3 at 2:05
add a comment |
$begingroup$
I see how the general case might vary. As for 𝐄(𝐴^2), in general, would you just expand the square of A and then see if you get something you can work with?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:59
$begingroup$
In accordance to the given information, a good start point would be to notice that $$textbf{Var}(A) = textbf{E}(A^{2}) - textbf{E}(A)^{2}$$ Once we know the value of $textbf{Var}(A)$ and $textbf{E}(A)$ can be easily calculated, you obtain the value of $textbf{E}(A^{2})$. More precisely, we have $$textbf{E}(A) = textbf{E}(aX+bY) = atextbf{E}(X) + btextbf{E}(Y) = amu_{1} + bmu_{2}$$
$endgroup$
– APC89
Feb 3 at 2:05
$begingroup$
I see how the general case might vary. As for 𝐄(𝐴^2), in general, would you just expand the square of A and then see if you get something you can work with?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:59
$begingroup$
I see how the general case might vary. As for 𝐄(𝐴^2), in general, would you just expand the square of A and then see if you get something you can work with?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:59
$begingroup$
In accordance to the given information, a good start point would be to notice that $$textbf{Var}(A) = textbf{E}(A^{2}) - textbf{E}(A)^{2}$$ Once we know the value of $textbf{Var}(A)$ and $textbf{E}(A)$ can be easily calculated, you obtain the value of $textbf{E}(A^{2})$. More precisely, we have $$textbf{E}(A) = textbf{E}(aX+bY) = atextbf{E}(X) + btextbf{E}(Y) = amu_{1} + bmu_{2}$$
$endgroup$
– APC89
Feb 3 at 2:05
$begingroup$
In accordance to the given information, a good start point would be to notice that $$textbf{Var}(A) = textbf{E}(A^{2}) - textbf{E}(A)^{2}$$ Once we know the value of $textbf{Var}(A)$ and $textbf{E}(A)$ can be easily calculated, you obtain the value of $textbf{E}(A^{2})$. More precisely, we have $$textbf{E}(A) = textbf{E}(aX+bY) = atextbf{E}(X) + btextbf{E}(Y) = amu_{1} + bmu_{2}$$
$endgroup$
– APC89
Feb 3 at 2:05
add a comment |
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$begingroup$
Concerning the calculation of the variance of a sum of random variables, search for "Bienaymé formula". Concerning the interpretation: What exactly are you thinking of? Do you have a general understanding of what the variance of a random variable is? Why should it be different for a sum? Concerning generelizations: Any formula you know for $E(f(X))$ (e.g. if you know a density) can be applied to calculate the variance. Is this what you have in mind?
$endgroup$
– Mars Plastic
Feb 3 at 1:21
$begingroup$
But in general, you cannot expect to calculate the variance of $X+Y$ just from the variances of $X$ and $Y$. If for example $X=Y$, we have $Var(X+Y)=4Var(X)$, but if they have the same distribution but are independent, we have $Var(X+Y)=2Var(X)$
$endgroup$
– Mars Plastic
Feb 3 at 1:25
$begingroup$
I guess I was not paying attention to the role of independence in how the sum might vary. Taking independence into account, I think I now see how the meaning of a variance of a sum not that different. Can you direct me to a link to learn more about expectations of functions of random variables?
$endgroup$
– Kumail Alhamoud
Feb 3 at 1:56