We know that the random variable $Z = \sum_i X_i$ formed by summing i.i.d $X_i \sim N(\mu, \sigma^2)$ with itself $n$ times is $Z \sim N(n\mu, n\sigma^2)$.

Given this knowledge, is it possible to take $m$ samples of $X$ by taking $m$ samples of $Z$, then dividing the samples by $n$? That is, suppose we have $m$ samples $(z_1, z_2, \dots, z_m)$ of $Z$, is $(\frac{z_1}{n}, \dots, \frac{z_m}{n})$ a set of $m$ samples from $X \sim N(\mu, \sigma^2)$?

  • 2
    $\begingroup$ $nX = Z \sim \mathcal{N}\left(n\mu,n^2\sigma^2\right)$, did you mean $\sum_{i=1}^n X_i =Y \sim \mathcal{N}\left(n\mu,n\sigma^2\right)$ where $X_i$ are i.i.d.? $\endgroup$ – Jonathan Lisic Nov 10 '16 at 6:49
  • $\begingroup$ sorry, the latter is what I meant, will edit. $\endgroup$ – Alex Nov 10 '16 at 22:50

Yes, but I'm not sure if that is what you really want. When you say adding $X$ to itself, that is just scaling by an integer, $nX_j = Z_j \sim \mathcal{N}\left(n\mu,n^2\sigma^2\right)$. By dividing by $n>0$ you are just scaling by the inverse of the original scalar. So, if $\left\{Z_1,...,Z_j,...,Z_m\right\}$ is an i.i.d. sample from the above distribution, then $\sum_{j=1}^m Z_j/n = \sum_{j=1}^m X_j\sim \mathcal{N}\left(m\mu,m\sigma^2\right)$.


Yes, and we can see this by considering the inverse transform sampling method on $N(0,1)$, with cumulative distribution function $F(x)$. The key point is that $N(\mu, \sigma^2)$ has CDF $G = F(\frac{x - \mu}{\sigma})$, that is, it is the CDF of $N(0,1)$, shifted and rescaled.

For a sample $u \sim U(0, 1)$ drawn from the uniform distribution, let $x = F^{-1}(u)$ be the sample drawn from $N(0, 1)$ using the inverse transform sampling method. The rescaled value $\sigma x + \mu$ is equal to $G^{-1}(u)$, thus can be taken as a sample from $N(\mu, \sigma^2)$.

The answer to:

suppose we have $m$ samples $(z_1, z_2, \dots, z_m)$ of $Z$, is $(\frac{z_1}{n}, \dots, \frac{z_m}{n})$ a set of $m$ samples from $X \sim N(\mu, \sigma^2)?$

is yes as well. Since,

$$x \sim N(0, 1) \Rightarrow \sigma x + \mu \sim N(\mu, \sigma^2)$$ and $$x \sim N(0, 1) \Rightarrow n\sigma x + n\mu \sim N(n\mu, n\sigma^2),$$

a sample $x_1 \sim N(\mu, \sigma^2)$ can be obtained from $x_2 \sim N(n\mu, n\sigma^2)$ by composition $f \circ g^{-1}$ of the functions:

$$f(x) = \sigma x + \mu$$ $$g(x) = n\sigma x + n\mu$$

such that

$$x_1 = \frac{\sigma(x_2 - n \mu)}{n\sigma} + \mu = \frac{x_2}{n}$$


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