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If I am sampling randomly from only the -sigma to +sigma interval of a normal distribution and rejecting all other numbers, does it imply that the probability density changes? If so, by what degree?


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If you are taking random samples, what do you mean by occurence? – Daniel Littlewood Feb 27 '14 at 22:20
@DanielLittlewood The probability density essentially, of each of those random samples – user131983 Feb 27 '14 at 22:24
up vote 1 down vote accepted

Let $f(x)$ be the density function of the original normal $X$.

The probability that you "keep" a number is the probability that the number obtained is between $\mu-\sigma$ and $\mu+\sigma$. This is approximately $0.6826$.

The resulting truncated distribution $Y$ has density function which is $0$ outside the interval $[\mu-\sigma,\mu+\sigma$. Inside the interval, it has density function $\frac{f(y)}{0.6826}$ (the $0.6826$ is approximate).

For $y$ between $\mu-\sigma$ and $\mu+\sigma$, the probability that $Y\le y$ is given by $$\Pr(Y\le y)=\frac{\Pr(X\le y)}{0.6826}.\tag{1}$$

Remark: To obtain Formula (1), let $A$ be the event $X\le y$, and let $B$ be the event $mu-\sigma \le X\le \mu+\sigma$. Then $$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$ We have $\Pr(B)\approx 0.6826$, and if $y$ is between $\mu-\sigma$ and $\mu+\sigma$, then $\Pr(A\cap B)=\Pr(X\le y)$.

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I would also mention the use of Bayes' formula here (unless you didn't use that, but it's the only way I can think of). EDIT: The connection is much more obvious after your edit. – Daniel Littlewood Feb 27 '14 at 22:30
I don't think of it as Bayes's Formula, more the definition of conditional probability. You are right, I should at least mention a bit of theory. – André Nicolas Feb 27 '14 at 22:34
@AndréNicolas Thanks. I guess I would need to multiply my original PDF with 1/0.6826. However, do you think a better way of choosing numbers from this interval would be to use Rejection sampling – user131983 Feb 27 '14 at 23:12
Maybe. For the normal one has to compute a bit, so a comparative "cost" analysis is not obvious. – André Nicolas Feb 27 '14 at 23:14
@AndréNicolas So, it isn't obvious if one method is slightly more accurate than the other then? – user131983 Feb 27 '14 at 23:17

Yes, because you can't choose numbers not within 1 standard deviation, the numbers you can choose have higher probabilities. The probability of numbers within the interval is the same, just scaled by $1/0.623$, or the probability a number is within 1 standard deviation.

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Thanks! So, if I wanted to know the probability of occurrence of a number in terms of the original distribution, being chosen from the truncated distribution, I would multiply my truncated distributions PDF with 0.623? – user131983 Feb 27 '14 at 23:22

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