Generate $r_i$ following condition $\sum_{i=1}^n \left|\frac{r_i}{\sigma_i}\right|^2\leq\chi^2_{n,\alpha}$ How can I generate $r_i$ for $1 \leq i \leq n$, such that $\sum_{i=1}^n \left|\frac{r_i}{\sigma_i}\right|^2\leq\chi^2_{n,\alpha}$, where  $\sigma_i^2$ is the variance of $r_i$ and $\chi^2_{n,\alpha}$ is a chi-squared value for $n$ degree of freedom and an $\alpha$ confidence level.
I truly appreciate your insights.
 A: Chi-squared distribution is defined in terms of normally distributed random variables. If $Z_1,\dots,Z_k$ are i.i.d. standard normal variables, then $ \sum_{i=1}^k Z_i^2 \sim \chi^2_k $. So to draw from chi-squared distribution with $k$ degrees of freedom, can use $k$ values drawn from standard normal. Alternatively, as in your case, you can draw $X_1,\dots,X_k$ from normal distribution with mean $0$ and standard deviation $\sigma^2$ and then take $Z_i = X_i / \sigma^2$ (so are you sure that you want to divide by variance..?).
set.seed(123)

f <- function() {
  n <- 20
  sigma <- 5
  r <- rnorm(n, 0, sigma)
  sum((r/sigma)^2)
}

x <- replicate(5000, f())
xx <- seq(0, 75, by = 0.01)

hist(x, 100, freq = FALSE)
lines(xx, dchisq(xx, df = n), col = "red")


If you need $100\alpha \% $ middle values, and $\alpha$ is big (say $0.95$), than the easiest way to go is to make multiple draws and then discard the draws that fall beyond the $100\alpha \% $ interval.
A: In general, as $n \rightarrow \infty$, then $(\chi_n ^2 -n)/2n \rightarrow N(0,1) $ for $ \chi^2$. Now we can establish the function $\Sigma_{i=1} ^n (\frac {r_i}{\sigma_i})^2 \leq \chi^2 _{n,\alpha} $ is true for the normal distribution $R$ for $r_i$. This is kind of actually a normal to chi-square transformation, and what you can do to generate the $r_i$ is usually select $\sigma_i^2 = 1$ for all $i$ and then you can find the $r_i^2$ values that correspond to the $\chi^2_n$ distribution. 
To consider $\alpha$ in this problem is really to determine if the value samples falls within the $1-\alpha$ confidence interval of $\chi^2_n$.
@Jolfaei The $r_i$  distribution is sampled from a normal distribution of mean $E(r_i)$ and variance $Var(r_i)$. It should come from $N(E(r_i), Var(r_i))$. If you know $n$, then you can sample from any normal distribution for a defined $E(r_i)$ and $Var(r_i)$. So lets say for example, you want to find the $\chi^2 _5$. You can actually pick the normal distribution parameters for each iteration. It should work because the tail end $\chi^2$ p-value increases as $n\rightarrow infty$ and by definition the $chi^2$ distribution is the sum of normal distributions.
