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Jul 9, 2021 at 0:56 comment added jld @CJR looks like you’ve got it but I’m happy to add any extra details if there’s any extra points that you’d like clarification on!
Jul 8, 2021 at 23:32 vote accept CJR
Jul 8, 2021 at 21:59 comment added CJR Thanks for expanding. I am still not quite sure how to calculate this expected value given the information I have (the weights $c_i$, the mean vector $\mu_{mix}$, the mean vector for each bi-var normal $\mu_i$, and the var-cov matrix for each bi-var normal $\Sigma_i$) . With that information, how do I calculate this expected value? Thank you! @jld
Jul 8, 2021 at 21:45 comment added jld @CJR yeah sure. I'm using the result that for a random vector $S$ with mean vector $\mu$ the variance is $\text{Var}[S] = \text E[SS^T] - \mu\mu^T$. We know $\mu$ so the only remaining thing is the first term of $\text E[SS^T]$ and that's just the expected value of the 2x2 matrices $ss^T$ weighted by the mixture density
Jul 8, 2021 at 21:01 comment added CJR Thanks for updating the post. I am still but unclear how the last line produces the variance-covariance matrix for the resulting mixture bivariate normal? If you could clarify that a bit more I'd really appreciate it. @jld
Jul 8, 2021 at 20:16 comment added jld @CJR i just updated for the mixture case since it seems like maybe that's what you meant?
Jul 8, 2021 at 20:16 history edited jld CC BY-SA 4.0
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Jul 8, 2021 at 19:30 comment added jld @CJR for sure! glad this helped
Jul 8, 2021 at 19:30 history edited jld CC BY-SA 4.0
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Jul 8, 2021 at 19:24 comment added CJR Thanks! This was my intuition & I appreciate the quick response.
Jul 8, 2021 at 19:08 history answered jld CC BY-SA 4.0