Sampling from a Normal Distribution 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?
Thanks
 A: 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 $\Pr(A\cap B)=\Pr(X\le y)$.
A: 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.
