Timeline for Normal Distribution
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2015 at 17:36 | comment | added | yoki | @Alia, the central limit theorem says that the distribution of an average of i.i.d variables converges to the Gaussian distribution for infinite $n$. So since these $Z_i=X_iI_i$ are indeed i.i.d variables, the CLT theorem applies. You would need to calculate the mean and variance to get the limiting distribution parameters. | |
| Jul 29, 2015 at 14:05 | comment | added | Alia | Guys, thanks a lot for your comments. I really approciate your help... I understood why the way I was thinking is not correct... But...Is there any way to show that $1/n*\sum X_{i]I_{i}$ is at least asymptotically normally distributed? ....maybe whith the central limit theorem? | |
| Jul 29, 2015 at 9:48 | comment | added | yoki | @Alia , the probability for each $X_i$ to be higher than its mean is $0.5$. The overall sum will be zero if each of the $X_i$ is higher than its mean, and the probability for this event is neither zero nor one (unless they exhibit specific types of dependencies), so it cannot be a Gaussian variable. | |
| Jul 29, 2015 at 9:40 | comment | added | Glen_b | The indicator is a random variable, as yoki said. It's $\text{Bernoulli}(\frac12)$ distributed. Further, $X_iI_i$ is clearly not normal, but a flipped, shifted half-nomal + a spike at $0$. $\sum X_iI_i$ is therefore not normal. | |
| Jul 29, 2015 at 9:16 | comment | added | Alia | $I$ is just an indicator... The general statement is actually: $X_{1},\ldots X_{n} \sim N(\mu, \sigma^{2})$ and I_{i}^{*}= \begin{cases} 1 & \text{if} X_{i}<\mu\\ \frac{1}{2} &\text{if} X_{i}=\mu\\ 0 & \text{if} X_{i} > \mu \end{cases} And I have to derive that $\frac{1}{n}\sum\limits_{i=1}^{n}X_{i}I_{i}$ is normally distributed. (The statement is in one book and I have to write down the proof) That's why $XI$ should be normally distributed as well.....I wanted to proof it for the entire sum with the induction... | |
| Jul 29, 2015 at 9:00 | comment | added | yoki | As far as I can see, $I$ is a random variable, and the question is whether or not the variable $Z=X\cdot I$ is a normal random variable, which it is not. (EDIT: this is a response to a now deleted comment...) | |
| Jul 29, 2015 at 8:41 | history | answered | yoki | CC BY-SA 3.0 |