Distribution of Sum of Sample Mean and Sample Variance from a Normal Population. Let $X_i\sim^{iid} N(\mu, \sigma^2)$. Let $\bar X$ and $S^2$ denote the usual, resp, sample mean and sample variance.
What is the distribution of $\bar X+S^2$?
Since we know that the sample mean and sample variance are independent in a Normal population, I guess my question would be equivalent to asking what is the resulting distribution of independent Normal distribution and a Chi-Squared distribution.
 A: Let $n$ be the sample size. Since $S^2$ is calculated from the sample generated by the $X_i$ we know that
$$X_i \sim N(\mu, \sigma^2)$$
$$\frac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)$$
The distribution of the sum of a Gaussian rv and a Chi-Squared rv is an instance of the Generalized Chi-Squared Distribution. A variable $\xi$ with the Generalized Chi-Squared Distribution can be defined as follows:
$$\xi = x +\sum_1^n w_i y_i \text { where } x\sim N(m,s),\;\; w_i \in \mathbb{R},\;\;y_i \sim \chi'^2(k_i,\lambda_i)\text{ and } y_i \text { independent}$$
Note that $\chi'^2(k_i,\lambda_i)$ is the non-central Chi-squared distribution, it is related to the Chi-squared as follows:
$$\chi^2(n) = \chi'^2(n,0)$$
In your case, we want the sum of a single Chi-Squared variable and a Normal, where the Chi-Squared is based on the sample variance of the sample from the Normal random variable:
$$\xi = x + wy \;\;\text{   where   } x\sim N\left(\mu,\frac{\sigma^2}{n}\right), \;y \sim \chi'^2(n-1,0)$$
What about the weight variable $w$? If $w=1$ then $\xi$ represents the following sum:
$$\bar{X} + \frac{(n-1)S^2}{\sigma^2}$$
But we want:
$$\bar{X} + S^2$$
Therefore, to get to this sum, we need to alter the weight applied to the chi-squared variable $y$:
$$w = \frac{\sigma^2}{n-1}$$
With this we have what we need:
$$\bar{X} + S^2 \sim \xi = x + wy \;\;\\\text{   where   } x\sim N\left(\mu,\frac{\sigma^2}{n}\right), \;y \sim \chi'^2(n-1,0),\;w = \frac{\sigma^2}{n-1}$$
In summary
$$\bar{X} + S^2 \sim \tilde{\chi}^2\left(\frac{\sigma^2}{n-1},n-1,0,\mu,\sigma\right) $$
