The covariance of a gaussian random variable and another one in its variance linear combination

Asked Nov 29 '20 at 00:05

Active Nov 29 '20 at 16:01

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Assume $y$ is a Gaussian random variable with $\text{Var}(y)= \sigma_y^2=\sum_i a_i^2\sigma_i^2$, where $\sigma_i$ corresponds to $x_i$, all $x_i$ are independent. $\mu_y = \sum a_i\mu_i$. I want to derive a formula for $\text{Cov}(y,x_i)$. I have tried the following:

$\text{Cov}(y,x_i) = \mathbb{E}[x_iy] - \mathbb{E}[x_i]\mathbb{E}[y]$. The last term is trivial to get. $\mathbb{E}[x_iy] \approx \mathbb{E}_z[\int \exp[-.5 (((x - \mu_i)^2)/\sigma_i ^2 + ((z/x - \mu_y)^2)/\sigma_y^2 )]/|x| dx]$, which is untractable I am afraind. So, I am wondering if there is a way to tackle this problem?

asked Nov 29 '20 at 00:05

rando

1

it looks like $y_{i} = \sum_{i} a_{i} x_{i}$. So, $cov(y, x_{i}) = cov(\sum_{i} a_{i} x_{i}, x_{i} ) = a_{i}$. – mlofton Nov 29 '20 at 05:20
@mlofton Yes, that is true. It seems I overcomplicated the problem – rando Nov 29 '20 at 15:37
@mlofton I think the covariance should be $a_i\sigma_i^2$ though. – rando Nov 29 '20 at 16:30
good catch. what i wrote was the correlation. my bad. – mlofton Nov 30 '20 at 15:19

The covariance of a gaussian random variable and another one in its variance linear combination

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