Timeline for Law of large numbers for transformed and non-transformed random variables
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 26, 2022 at 19:55 | comment | added | Aditya Mehrotra | @whuber This clears everything up, thank you! | |
| Jul 26, 2022 at 19:12 | comment | added | whuber♦ | It's even simpler: when the probability law of $X$ is $p,$ $E_p[f(X)]$ is the expectation, or "true mean," of $f(X).$ See stats.stackexchange.com/questions/467961 for one explanation or search our site for LOTUS for more. | |
| Jul 26, 2022 at 16:12 | comment | added | Aditya Mehrotra | @whuber So in that case, is it true that $E_{p(x)} \{f(X)\}$ is the true mean $\mu$ of $f(X)$ iff $f(X)$ has a PDF and $p(x)$ is that PDF? | |
| Jul 26, 2022 at 15:50 | comment | added | whuber♦ | As @GordonSmyth pointed out, this makes little sense. $X$ is the random variable. All expectations are taken with respect to its distribution. In fact, even when $X$ has a PDF, $f(X)$ might not have one. As an example, let $f(X)$ be the sign of $X:$ now $f(X)$ is a binary variable. | |
| Jul 26, 2022 at 14:47 | comment | added | Aditya Mehrotra | @whuber I see, so the expectation being considered in the case with $f(\cdot)$ is not the expectation of $f(X)$ w.r.t its own pdf but rather the expectation of $f(X)$ under $p(x)$. If I had omitted the $p(x)$ subscript and just had $\mathbb{E}\{f(X)\}$, would this imply that I would have an expectation under the transformed PDF of $X$ that I talk about in the question? | |
| Jul 26, 2022 at 13:17 | comment | added | whuber♦ | The notation is awful but it's relatively clear. (One has to make assumptions about how "$X_i,$" "$X,$" and "$x$" might be related and the specific meaning of $p$ is not entirely evident). However, in the statement you are asking about, the only random variable in sight is $X,$ whence "$\mathbb{E}_{p(x)}$" must refer to expectation relative to $X.$ | |
| Jul 26, 2022 at 13:05 | history | edited | kjetil b halvorsen♦ |
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| Jul 26, 2022 at 7:38 | comment | added | Gordon Smyth | $p(x)$ is simply the density of $X$. This remains true both in the original LLN statement at the start of your question and in the generalized version with $f(X_i)$ replacing $X_i$. There is no need in the theorem to transform the pdf. | |
| Jul 26, 2022 at 7:22 | history | edited | Aditya Mehrotra | CC BY-SA 4.0 |
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| Jul 26, 2022 at 7:17 | history | asked | Aditya Mehrotra | CC BY-SA 4.0 |