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Setup

Suppose I have two normally distributed random variables, $s_\mathrm{prev}$ (with mean $\mu_\mathrm{prev}$ and variance $\sigma^2_\mathrm{prev}$) and $s_\mathrm{curr}$ (with mean $\mu_\mathrm{curr}$ and variance $\sigma^2_\mathrm{curr}$). Now suppose I combine them as follows:

$r_\mathrm{curr} = \omega_\mathrm{prev}s_\mathrm{prev}+(1-\omega_\mathrm{prev})s_\mathrm{curr}$,

where $\omega_\mathrm{prev}$ lies between 0 and 1.

The mean of $r_\mathrm{curr}$ is $\omega_\mathrm{prev}\mu_\mathrm{prev} + (1-\omega_\mathrm{prev})\mu_\mathrm{curr}$.

Question

What is the variance of $r_\mathrm{curr}$?

According to the answers here, the variance is $\omega_\mathrm{prev}\sigma^2_\mathrm{prev}+(1-\omega_\mathrm{prev})\sigma^2_\mathrm{curr}+\omega_\mathrm{prev}(1-\omega_\mathrm{prev})(\mu_\mathrm{prev}-\mu_\mathrm{curr})^2$.

But these answers refer to a mixture of two normal pdfs, not random variables.

I've also seen it said that the variance I'm looking for is $\omega^2_\mathrm{prev}\sigma^2_\mathrm{prev}+(1-\omega_\mathrm{prev})^2\sigma^2_\mathrm{curr}$. If this is correct, how is this derived?

abcd
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    Your formula for the variance does not seem right. The question you are referring to is about a linear combination of pdf's, while yours is about a linear combination of random variables, which are different objects. Moreover, what is your definition of the RMSE? It usually implies samples or observations. – Camille Gontier Jun 12 '23 at 20:28
  • @CamilleGontier i'll update the question – abcd Jun 12 '23 at 20:39
  • The question about the linear combination is a FAQ, answered at https://stats.stackexchange.com/questions/38721/covariance-of-a-variable-and-a-linear-combination-of-other-variables/38722#38722. – whuber Jun 12 '23 at 21:13

1 Answers1

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Your result ($\omega^2_{prev} \sigma^2_{prev} + (1-\sigma_{prev})^2 \sigma^2_{curr}$) is correct if $s_{prev}$ and $s_{curr}$ are independent. In this case, the result naturally follows from the properties of the variance when applied to a linear combination of random variables, since $Cov(s_{prev},s_{curr}) = 0$.

Camille Gontier
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  • This answer evidently includes some typographical errors. Another issue is that the original post asks two different questions: one is about a mixture and the other is about a linear combination -- and both have been answered (many times) in other threads here on CV. – whuber Jun 12 '23 at 21:12