Solved – Sum of sample mean and sample variance sampling distribution

distributionsprobability

Let $X_1, X_2, \cdots, X_n$ be an identical and independently distributed sample from $N(\mu, \sigma^2)$, define:

$$D = \frac{1}{t}\left[\overline{X} + \frac{1-\rho}{2} S^2\right]$$

where:

$t$ and $p$ are fixed constants

$\overline{X} = \frac{1}{n}\sum_{i=1}^n X_i$, i.e, the sample mean

$S^2 = \frac{1}{n} \sum_{i=1}^n \left(X_i-\overline{X}\right)^2$, i.e, the biased sample variance

Then what is the sampling distribution of $D$?

I know that $\overline{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right)$ and $n\frac{S^2}{\sigma^2} \sim \chi^2(n-1)$, but how do I derive the sampling distribution of $D$?

Best Answer

I would go with Monte Carlo simulation. Pretty straightforward and less chance of mistakes. There may be some slick math-stats solution too, but Monte Carlo is probably faster.

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