How can I understand the mean of convoluted Gaussian function is $mu_1 + mu_2$. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How can I make some progress on this Gaussian-looking integral?Integrating Normal Distribution with $1$, $x$, and $x^2$Functions with closed-form expectations under Gaussian PDFImpact of gaussian mean shift on top end distributionIntegral of the product of Gaussian PDF and Q function involving natural logGaussian Integrals and Grassmann AlgebraRelation between Taylor Expansion and multivariate Gaussian Distributionfinding area under the curve of a valueIs there an analytic form for the squared error of the difference of two univariate Gaussians?Kalman filter: the bayesian approach derivation some clarifications
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How can I understand the mean of convoluted Gaussian function is $mu_1 + mu_2$.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How can I make some progress on this Gaussian-looking integral?Integrating Normal Distribution with $1$, $x$, and $x^2$Functions with closed-form expectations under Gaussian PDFImpact of gaussian mean shift on top end distributionIntegral of the product of Gaussian PDF and Q function involving natural logGaussian Integrals and Grassmann AlgebraRelation between Taylor Expansion and multivariate Gaussian Distributionfinding area under the curve of a valueIs there an analytic form for the squared error of the difference of two univariate Gaussians?Kalman filter: the bayesian approach derivation some clarifications
$begingroup$
I struggle to understand the mean of convoluted two Gaussian functions is $mu_1 + mu_2$ instead of $(mu_1 + mu_2)/2$. Could someone provide visual explaination?
gaussian-integral
asked Mar 27 at 19:57
Alex GaoAlex Gao
1125
$endgroup$
add a comment |
$begingroup$
I struggle to understand the mean of convoluted two Gaussian functions is $mu_1 + mu_2$ instead of $(mu_1 + mu_2)/2$. Could someone provide visual explaination?
gaussian-integral
asked Mar 27 at 19:57
Alex GaoAlex Gao
1125
$endgroup$
$begingroup$
Can you explain in your question why you think it should be halved? Do you understand that if $X$ and $Y$ have distributions $f$ and $g$ then $X + Y$ has distribution $f * g$?
$endgroup$
– Trevor Gunn
Mar 27 at 20:13
add a comment |
$begingroup$
I struggle to understand the mean of convoluted two Gaussian functions is $mu_1 + mu_2$ instead of $(mu_1 + mu_2)/2$. Could someone provide visual explaination?
gaussian-integral
asked Mar 27 at 19:57
Alex GaoAlex Gao
1125
$endgroup$
I struggle to understand the mean of convoluted two Gaussian functions is $mu_1 + mu_2$ instead of $(mu_1 + mu_2)/2$. Could someone provide visual explaination?
gaussian-integral
gaussian-integral
asked Mar 27 at 19:57
Alex GaoAlex Gao
1125
asked Mar 27 at 19:57
Alex GaoAlex Gao
1125
edited Mar 27 at 20:03
Alex Gao
asked Mar 27 at 19:57
Alex GaoAlex Gao
1125
asked Mar 27 at 19:57
Alex GaoAlex Gao
1125
asked Mar 27 at 19:57
Alex GaoAlex Gao
1125
1125
$begingroup$
Can you explain in your question why you think it should be halved? Do you understand that if $X$ and $Y$ have distributions $f$ and $g$ then $X + Y$ has distribution $f * g$?
$endgroup$
– Trevor Gunn
Mar 27 at 20:13
add a comment |
$begingroup$
Can you explain in your question why you think it should be halved? Do you understand that if $X$ and $Y$ have distributions $f$ and $g$ then $X + Y$ has distribution $f * g$?
$endgroup$
– Trevor Gunn
Mar 27 at 20:13
$begingroup$
Can you explain in your question why you think it should be halved? Do you understand that if $X$ and $Y$ have distributions $f$ and $g$ then $X + Y$ has distribution $f * g$?
$endgroup$
– Trevor Gunn
Mar 27 at 20:13
$begingroup$
Can you explain in your question why you think it should be halved? Do you understand that if $X$ and $Y$ have distributions $f$ and $g$ then $X + Y$ has distribution $f * g$?
$endgroup$
– Trevor Gunn
Mar 27 at 20:13
add a comment |
1 Answer
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$begingroup$
If $X$ and $Y$ are two independent continuous random variables with means $E[X]=mu_X$ and $E[Y] = mu_Y$, then the density of their sum $X+Y$ is the convolution of the densities of $X$ and $Y$. In the special case when $X$ and $Y$ are Gaussian random variables, $X+Y$ is also a Gaussian random variable. But, regardless of whether $X$ and $Y$ are Gaussian or not, independent or not, or continuous or not, it is true that
$$E[X+Y] = E[X] + E[Y].$$
This result is referred to as the linearity of expectation, and explains why the mean of $X+Y$ is $mu_X + mu_Y$, and not the arithmetic average $fracmu_X+mu_Y2$ that you are hoping to see.
Visually, the densities of independent Gaussian $X$ and $Y$ are bell-shaped curves centered at $mu_X$ and $mu_Y$ respectively, and the value of the convolution of these densities at $z$ (a.k.a the numerical value of the density of $X+Y$ at $z$) is obtained by
- "flipping" the density of $Y$ about the origin the result of which operation looks for all the world just like the density of $Y$ centered at $-mu_Y$ instead of at $mu_Y$.
- sliding the flipped density to the right by $z$
- multiplying the density of $X$ with the flipped and slid density of $Y$ and integrating the product from $-infty$ to $infty$.
If your brain has not already boggled at all these shenanigans and can still visualize a tad more, consider that the maximum value of the integral will occur when the two peaks align, that is, when the flipped density of $Y$ has been slid so that its peak at $-mu_Y$ has shifted due to the slide so that it coincides with the peak of the $X$ density which is at $mu_X$, i.e. a total slide of $mu_X+mu_Y$. So, the peak of the convolution is at $mu_X+mu_Y$ and so the mean of $X+Y$ is $mu_X+mu_Y$.
answered Mar 27 at 20:16
Dilip SarwateDilip Sarwate
19.5k13078
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add a comment |
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$begingroup$
If $X$ and $Y$ are two independent continuous random variables with means $E[X]=mu_X$ and $E[Y] = mu_Y$, then the density of their sum $X+Y$ is the convolution of the densities of $X$ and $Y$. In the special case when $X$ and $Y$ are Gaussian random variables, $X+Y$ is also a Gaussian random variable. But, regardless of whether $X$ and $Y$ are Gaussian or not, independent or not, or continuous or not, it is true that
$$E[X+Y] = E[X] + E[Y].$$
This result is referred to as the linearity of expectation, and explains why the mean of $X+Y$ is $mu_X + mu_Y$, and not the arithmetic average $fracmu_X+mu_Y2$ that you are hoping to see.
Visually, the densities of independent Gaussian $X$ and $Y$ are bell-shaped curves centered at $mu_X$ and $mu_Y$ respectively, and the value of the convolution of these densities at $z$ (a.k.a the numerical value of the density of $X+Y$ at $z$) is obtained by
- "flipping" the density of $Y$ about the origin the result of which operation looks for all the world just like the density of $Y$ centered at $-mu_Y$ instead of at $mu_Y$.
- sliding the flipped density to the right by $z$
- multiplying the density of $X$ with the flipped and slid density of $Y$ and integrating the product from $-infty$ to $infty$.
If your brain has not already boggled at all these shenanigans and can still visualize a tad more, consider that the maximum value of the integral will occur when the two peaks align, that is, when the flipped density of $Y$ has been slid so that its peak at $-mu_Y$ has shifted due to the slide so that it coincides with the peak of the $X$ density which is at $mu_X$, i.e. a total slide of $mu_X+mu_Y$. So, the peak of the convolution is at $mu_X+mu_Y$ and so the mean of $X+Y$ is $mu_X+mu_Y$.
answered Mar 27 at 20:16
Dilip SarwateDilip Sarwate
19.5k13078
$endgroup$
add a comment |
$begingroup$
If $X$ and $Y$ are two independent continuous random variables with means $E[X]=mu_X$ and $E[Y] = mu_Y$, then the density of their sum $X+Y$ is the convolution of the densities of $X$ and $Y$. In the special case when $X$ and $Y$ are Gaussian random variables, $X+Y$ is also a Gaussian random variable. But, regardless of whether $X$ and $Y$ are Gaussian or not, independent or not, or continuous or not, it is true that
$$E[X+Y] = E[X] + E[Y].$$
This result is referred to as the linearity of expectation, and explains why the mean of $X+Y$ is $mu_X + mu_Y$, and not the arithmetic average $fracmu_X+mu_Y2$ that you are hoping to see.
Visually, the densities of independent Gaussian $X$ and $Y$ are bell-shaped curves centered at $mu_X$ and $mu_Y$ respectively, and the value of the convolution of these densities at $z$ (a.k.a the numerical value of the density of $X+Y$ at $z$) is obtained by
- "flipping" the density of $Y$ about the origin the result of which operation looks for all the world just like the density of $Y$ centered at $-mu_Y$ instead of at $mu_Y$.
- sliding the flipped density to the right by $z$
- multiplying the density of $X$ with the flipped and slid density of $Y$ and integrating the product from $-infty$ to $infty$.
If your brain has not already boggled at all these shenanigans and can still visualize a tad more, consider that the maximum value of the integral will occur when the two peaks align, that is, when the flipped density of $Y$ has been slid so that its peak at $-mu_Y$ has shifted due to the slide so that it coincides with the peak of the $X$ density which is at $mu_X$, i.e. a total slide of $mu_X+mu_Y$. So, the peak of the convolution is at $mu_X+mu_Y$ and so the mean of $X+Y$ is $mu_X+mu_Y$.
answered Mar 27 at 20:16
Dilip SarwateDilip Sarwate
19.5k13078
$endgroup$
add a comment |
$begingroup$
If $X$ and $Y$ are two independent continuous random variables with means $E[X]=mu_X$ and $E[Y] = mu_Y$, then the density of their sum $X+Y$ is the convolution of the densities of $X$ and $Y$. In the special case when $X$ and $Y$ are Gaussian random variables, $X+Y$ is also a Gaussian random variable. But, regardless of whether $X$ and $Y$ are Gaussian or not, independent or not, or continuous or not, it is true that
$$E[X+Y] = E[X] + E[Y].$$
This result is referred to as the linearity of expectation, and explains why the mean of $X+Y$ is $mu_X + mu_Y$, and not the arithmetic average $fracmu_X+mu_Y2$ that you are hoping to see.
Visually, the densities of independent Gaussian $X$ and $Y$ are bell-shaped curves centered at $mu_X$ and $mu_Y$ respectively, and the value of the convolution of these densities at $z$ (a.k.a the numerical value of the density of $X+Y$ at $z$) is obtained by
- "flipping" the density of $Y$ about the origin the result of which operation looks for all the world just like the density of $Y$ centered at $-mu_Y$ instead of at $mu_Y$.
- sliding the flipped density to the right by $z$
- multiplying the density of $X$ with the flipped and slid density of $Y$ and integrating the product from $-infty$ to $infty$.
If your brain has not already boggled at all these shenanigans and can still visualize a tad more, consider that the maximum value of the integral will occur when the two peaks align, that is, when the flipped density of $Y$ has been slid so that its peak at $-mu_Y$ has shifted due to the slide so that it coincides with the peak of the $X$ density which is at $mu_X$, i.e. a total slide of $mu_X+mu_Y$. So, the peak of the convolution is at $mu_X+mu_Y$ and so the mean of $X+Y$ is $mu_X+mu_Y$.
answered Mar 27 at 20:16
Dilip SarwateDilip Sarwate
19.5k13078
$endgroup$
If $X$ and $Y$ are two independent continuous random variables with means $E[X]=mu_X$ and $E[Y] = mu_Y$, then the density of their sum $X+Y$ is the convolution of the densities of $X$ and $Y$. In the special case when $X$ and $Y$ are Gaussian random variables, $X+Y$ is also a Gaussian random variable. But, regardless of whether $X$ and $Y$ are Gaussian or not, independent or not, or continuous or not, it is true that
$$E[X+Y] = E[X] + E[Y].$$
This result is referred to as the linearity of expectation, and explains why the mean of $X+Y$ is $mu_X + mu_Y$, and not the arithmetic average $fracmu_X+mu_Y2$ that you are hoping to see.
Visually, the densities of independent Gaussian $X$ and $Y$ are bell-shaped curves centered at $mu_X$ and $mu_Y$ respectively, and the value of the convolution of these densities at $z$ (a.k.a the numerical value of the density of $X+Y$ at $z$) is obtained by
- "flipping" the density of $Y$ about the origin the result of which operation looks for all the world just like the density of $Y$ centered at $-mu_Y$ instead of at $mu_Y$.
- sliding the flipped density to the right by $z$
- multiplying the density of $X$ with the flipped and slid density of $Y$ and integrating the product from $-infty$ to $infty$.
If your brain has not already boggled at all these shenanigans and can still visualize a tad more, consider that the maximum value of the integral will occur when the two peaks align, that is, when the flipped density of $Y$ has been slid so that its peak at $-mu_Y$ has shifted due to the slide so that it coincides with the peak of the $X$ density which is at $mu_X$, i.e. a total slide of $mu_X+mu_Y$. So, the peak of the convolution is at $mu_X+mu_Y$ and so the mean of $X+Y$ is $mu_X+mu_Y$.
answered Mar 27 at 20:16
Dilip SarwateDilip Sarwate
19.5k13078
edited Mar 27 at 20:33
answered Mar 27 at 20:16
Dilip SarwateDilip Sarwate
19.5k13078
answered Mar 27 at 20:16
Dilip SarwateDilip Sarwate
19.5k13078
answered Mar 27 at 20:16
Dilip SarwateDilip Sarwate
19.5k13078
19.5k13078
add a comment |
add a comment |
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$begingroup$
Can you explain in your question why you think it should be halved? Do you understand that if $X$ and $Y$ have distributions $f$ and $g$ then $X + Y$ has distribution $f * g$?
$endgroup$
– Trevor Gunn
Mar 27 at 20:13