Timeline for Estimating the average of two gaussians' mean with minimal squared error
Current License: CC BY-SA 4.0
13 events
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| Oct 22, 2021 at 9:20 | history | edited | R B | CC BY-SA 4.0 |
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| Oct 21, 2021 at 20:11 | comment | added | Michael Hardy | What you have said is also consistent with some scenarios in which the pair $(X,Y)$ is not jointly normally distributed; for example $$ \frac{X-\mu_1}{\sigma_1} = \pm\frac{Y-\mu_2}{\sigma_2} $$ where plus of minus is random and independent of $Y$ and the two signs have equal probabilities. $\qquad$ | |
| Oct 21, 2021 at 20:11 | comment | added | Michael Hardy | What you have said is also consistent with independence of $X$ and $Y$, which would yield a very different joint distribution. Intermediate between those two extremes is this: $$ \Pr\left( \frac{X_1-\mu_2}{\sigma_1} = \rho\cdot\frac{Y-\mu_2}{\sigma_2} + \sqrt{1-\rho^2}\cdot Z \right) = 1 $$ where $Z\sim \operatorname N(0,1)$ is independent of $Y,$ so that $X$ is determined by $Y$ and $Z$ and $\operatorname{cor}(X,Y) = \rho. \qquad$ | |
| Oct 21, 2021 at 20:10 | comment | added | Michael Hardy | You have said nothing of the joint distribution of $X$ and $Y.$ What you have said is consistent with $$ \Pr\left( \frac{X-\mu_1}{\sigma_1} = \frac{Y-\mu_2}{\sigma_2} \right) = 1. $$ | |
| Oct 21, 2021 at 20:09 | comment | added | Michael Hardy | Here I will assume you mean what is usually denoted by $X\sim\operatorname N(\mu_1,\sigma_1^2)$ and $Y\sim\operatorname N(\mu_2,\sigma_2^2)\,\ldots \quad$ | |
| Oct 21, 2021 at 16:50 | vote | accept | R B | ||
| Oct 21, 2021 at 14:13 | comment | added | user44143 | Even in that context the MLE estimate would be the average of X and Y, and I don’t see any other principled estimation method that would give anything different. | |
| Oct 21, 2021 at 13:50 | answer | added | Iosif Pinelis | timeline score: 1 | |
| Oct 21, 2021 at 13:49 | comment | added | R B | @Matt, The problem comes from trying to estimate the average of measurements taken by heterogeneous devices that have different bandwidths for transmitting their signal. This means that the devices have different quantization errors, which we can estimate as we know the number of bits they transmitted. To simplify the model, we can assume that the quantization noise is Gaussian, although this is not accurate (and depends on the quantization method). | |
| Oct 21, 2021 at 13:47 | comment | added | R B | Thanks, @IosifPinelis. Can you please explain why? Let's even say that $\sigma_1=0$; wouldn't it make more sense to have an estimate closer to $X$? | |
| Oct 21, 2021 at 13:39 | comment | added | user44143 | What is the setting and where does your intuition come from? I suspect that your interest here is not purely mathematical, and you’d probably get more insightful answers by providing more context. | |
| Oct 21, 2021 at 13:34 | comment | added | Iosif Pinelis | Without restrictions on $\mu_1,\mu_2$, I think you will not get the desired result. | |
| Oct 21, 2021 at 13:20 | history | asked | R B | CC BY-SA 4.0 |