Timeline for Sampling probability distribution-how many samples?
Current License: CC BY-SA 3.0
13 events
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| Mar 7, 2017 at 7:38 | comment | added | bissi | @JohnK. I don't get it. $P(−0.5≤F_n(k)−P(X≤k)≤0.5)= 2 \mathbb{N}(0, b(1-b)/N)[0.5]$ | |
| Mar 6, 2017 at 19:46 | history | edited | JohnK | CC BY-SA 3.0 |
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| Mar 6, 2017 at 18:47 | comment | added | JohnK | @bissi No it doesn't. Like I said in my previous comment if you replace the unknown $b = P(X\leq k)$ with $F_n(k)$ in the variance, you have a legitimate way of computing such probabilities. | |
| Mar 6, 2017 at 18:41 | comment | added | bissi | @JohnK This leads to a chicken and egg problem, as the probability that this difference is less than 0.5 is a normal distribution with mean 0 but the variance is unknown, as the $b$ is unknown in the first place... | |
| Mar 6, 2017 at 10:06 | comment | added | JohnK | @bissi No, the mean of this absolute value is not zero. The variable $F_n(k)-P(X\leq k)$ (without the absolute value) is normally distributed with mean zero and the variance you say. Based on this distribution you can compute the probability that this difference is, say less than 0.5 by $$ P\left( |F_n(k)-P(X\leq k)| \leq 0.5 \right) = P\left( -0.5 \leq F_n(k)-P(X\leq k) \leq 0.5 \right) $$ using the normal distribution. This is what you had in mind, right? | |
| Mar 6, 2017 at 9:58 | comment | added | bissi | @JohnK got it. So for completeness, the mean of this absolute value is 0 and the variance is actually simply $b(1-b)/n$ where $b$ is P(X\le k)$ | |
| Mar 5, 2017 at 19:37 | comment | added | JohnK | @bissi For the variance you use the estimate that you have for $P(X\leq k )$, namely $F_n(k)$. It can be shown that the asymptotic normal distribution still holds after you plug-in the estimate. $F_n(k)-P(X\leq k)$ has by definition mean zero, so you don't have to worry about that. Just find the density of the the absolute value of a normal variable. | |
| Mar 5, 2017 at 15:23 | comment | added | bissi | @JohnK how do I get the mean and variance of this normal distribution as I do not know the value $P(X \le k)$? | |
| Mar 3, 2017 at 20:05 | history | edited | JohnK | CC BY-SA 3.0 |
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| Mar 3, 2017 at 15:26 | comment | added | JohnK | @A.Webb The OP is interested in making probabilistic statements, the classical CLT is all that he needs for this. | |
| Mar 3, 2017 at 15:25 | comment | added | A. Webb | Asymptotically, yes, but it's the rate of convergence that's essential to the question, no? | |
| Mar 3, 2017 at 15:21 | history | edited | JohnK | CC BY-SA 3.0 |
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| Mar 3, 2017 at 15:14 | history | answered | JohnK | CC BY-SA 3.0 |