Timeline for Estimating a statistic by combining two different data sources
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jan 16, 2022 at 19:20 | comment | added | causative | Thanks, that makes sense. | |
| Jan 16, 2022 at 19:20 | vote | accept | causative | ||
| Jan 16, 2022 at 16:54 | comment | added | F. Tusell | If $X$ and $Y$ both have mean $\theta$, that you want to estimate, an unbiased estimator is of the form $\hat\theta = cX+(1-c)Y$, and has variance $c^2\sigma_X^2 + (1-c)^2\sigma_Y^2$. If you minimize this wrt $c$ you arrive to the quite intuitive result that the weights of the observations should be inversely proportional to their variances. This generalize to more than two observations. For a suitable $c$ you whill have a variance which is less than that obtainable from $X$ or $Y$ alone. | |
| Jan 15, 2022 at 18:57 | comment | added | causative | Can you point me to something about appropriately weighting the observations in the case when there is no significant difference in means? | |
| Jan 15, 2022 at 13:31 | comment | added | F. Tusell | @causative, yes, you can use a t-test if you do the right correction, thats why I wrote "rodinary t-test". What you say about the variances assumes you pool A and B observations with equal weights, something which you would never want to do. | |
| Jan 14, 2022 at 20:24 | comment | added | causative | You can still use a t-test when the variances are different, see stat.yale.edu/Courses/1997-98/101/meancomp.htm "Tests of Significance for Two Unknown Means and Unknown Standard Deviations." The logic of only using A when there's no significant difference is that the variance of B is very high compared to the variance of A, so if we pooled A with B before gathering enough observations, the variance of the overall estimate would be higher than if we just used A. | |
| Jan 14, 2022 at 5:47 | comment | added | F. Tusell | @causative, a ordinary t-test requires common variance in the two populations. I cannot follow the logic: if there were no significant difference, you would probably want to pool all observations for an unbiased estimator, not just use A. | |
| Jan 13, 2022 at 18:54 | comment | added | causative | What if I just did a t-test for difference of means between the A samples and the B samples? If there is no significant difference then my estimate could be the mean of just the A samples, and if there is a significant difference then my estimate could be the mean of just the B samples. I could also use the test statistic to gradually interpolate between the two cases. | |
| Jan 13, 2022 at 15:42 | history | edited | F. Tusell | CC BY-SA 4.0 |
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| Jan 13, 2022 at 15:41 | comment | added | F. Tusell | @clarivate, you do not need a beforehand estimation of the bias, you can just estimate $k$ with the other parameters. Time-invariance of $\theta$ can be accommodated setting the variance of $\theta_t$ to zero. | |
| Jan 13, 2022 at 15:11 | comment | added | causative | I should also clarify that $\theta$ is not time-varying. | |
| Jan 13, 2022 at 14:56 | comment | added | causative | The bias is assumed not too large, but is unknown. It can only be estimated by comparing the long run sample mean of A with the long run sample mean of B. And if you have enough observations from B to do this accurately, then you already have an accurate estimate of $\theta$ from B alone, and no longer need A. So estimating the bias of A is really most of the problem. | |
| Jan 13, 2022 at 10:58 | history | answered | F. Tusell | CC BY-SA 4.0 |