Timeline for combining a random variable and its inverse
Current License: CC BY-SA 3.0
14 events
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| Mar 4, 2015 at 14:55 | comment | added | Aksakal | @jobxyz1, Ok, the good thing is that it doesn't look like Gaussian, otyherwise you'd be in trouble, see this post, you wouldn't be able to invert the reading of Y meaningfully. Do you have any intuition about the nature of errors in Y? Its distribution or maybe the source? | |
| Mar 4, 2015 at 14:26 | comment | added | jobxyz1 | first var [0] 2.187 [1] -1.356 [2] 1.6815 [3] 1.07 [4] 1.2885 [5] 3.1345 [6] -0.1245 [7] 2.672 [8] 2.1955 [9] 1.2 [10] 3.9365 [11] 2.145 [12] 6.7495 second var [0] 0.1005 [1] -0.062 [2] 0.02775 [3] 0.03175 [4] 0.1525 [5] 0.40325 [6] 0.408 [7] 0.37425 [8] 0.31525 [9] 0.18475 [10] 0.2125 [11] 0.2985 [12] 0.249166667 [13] 0.06375 [14] 0.16525 [15] 0.3025 [16] 0.2575 [17] 0.1195 [18] 0.017 [19] 0.53225 [20] 0.48125 [21] 0.3005 [22] 0.4815 [23] 0.182333333 [24] 0.376 [25] -0.0425 [26] 0.24775 [27] 0.2635 [28] 0.04025 | |
| Mar 4, 2015 at 13:19 | history | edited | Aksakal | CC BY-SA 3.0 |
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| Mar 4, 2015 at 12:54 | comment | added | Aksakal | Maybe you have to deal with noise first. With stats you can do only so much. Can you show the data? | |
| Mar 4, 2015 at 12:50 | comment | added | jobxyz1 | Both variables are noisy... | |
| Mar 4, 2015 at 11:51 | comment | added | Aksakal | How about inverting X variable? Is it as noisy? | |
| Mar 4, 2015 at 10:30 | comment | added | jobxyz1 | Thank you both. Aksakal, your analogy of electrical measurements is spot-on. The problem that I have, though, is that the measurements are very noisy. So some of them show up negative, or close to 0 which creates a whole host of issues when using 1/x. So what I did is take the median and interquartile range of each variable, and that gets me reasonable results for each variable's "center of gravity" and dispersion, as I gave in the example. (not exactly mean and SD - but for my purposes - OK) The problem is in combining these medians. | |
| Mar 3, 2015 at 17:55 | comment | added | whuber♦ | I am not criticizing your assumptions (because I can't figure them out). I am only pointing out that what you are writing is inconsistent with the information in the question and likely is not saying what you think it is. Given that the question itself is inherently ambiguous, it is difficult to see how you can justify any answer at this time unless you are quite explicit and clear about what you think the data are and what you believe the question to be. | |
| Mar 3, 2015 at 17:52 | comment | added | Aksakal | @whuber, read my update in the post. No, I'm not assuming any pairing. The analogy is to measure the resistance in Ohms directly vs. in Amperes (given constant voltage). So, $\frac{10 V}{y A}\sim x\Omega$ | |
| Mar 3, 2015 at 17:49 | history | edited | Aksakal | CC BY-SA 3.0 |
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| Mar 3, 2015 at 17:44 | comment | added | whuber♦ | Starting at "$1/y_j = x_j$", which does not match anything you have written so far (because it implies that somehow the two datasets have been paired), your interpretation of the question and your assumptions become progressively less clear. | |
| Mar 3, 2015 at 17:41 | comment | added | Aksakal | @whuber, OP wrote that $x_i$ is 30 measurements and $y_j$ is 12 measurements. OP computed means $\bar x$ and $\bar y$. OP stated that underlying physics suggests that $\frac{1}{y_j}$ should correspond to $x$ somehow. I'm suggesting not to average $\bar y$, but to get an average of $\frac{1}{y_j}$ directly. | |
| Mar 3, 2015 at 16:56 | comment | added | whuber♦ | This answer is confusing. It is not the case that $E[1/y_j] = x_j$, as Jensen's inequality immediately shows. The whole point of the question is that $y_j$ and $x_j$ are measurements with error. In fact, there is not even a pairing of the $y_j$ and $x_i$: they are two (perhaps independent) datasets. | |
| Mar 3, 2015 at 16:47 | history | answered | Aksakal | CC BY-SA 3.0 |