I want to calculate the variance of sample mean of an iid random sample $X_{1}, X_{2}, \cdots X_{n}$ of the random variable $X$ which is distributed as $N(\mu, \sigma^{2})$ with probability $(1-\delta)$ and $f(x)$ with probability $\delta$.

The pdf $f(x)$ is any density with mean $\theta$ and variance $\tau^{2}$.

Statistical inference (2nd edition) by Casella and Berger says

$$Var(\bar{X}) = (1-\delta) \frac{\sigma^{2}}{n} + \delta \frac{\tau^{2}}{n} + \frac{\delta(1-\delta)(\theta - \mu)^{2}}{n}$$

but I cannot understand how I can arrive here. All I can guess is that since $\bar{X}$ is a sample mean, $Var(\bar{X}) = \frac{Var(X)}{n}$, but I cannot seem to calculate the variance of $X$ in the first place. Originally, I thought I can write $X$ as a linear combination of two random variables, $X = (1-\delta)Y_{1} + \delta Y_{2}$, with $Y_{1}$ distributed as $N(\mu, \sigma^{2})$ and $Y_{2}$ distributed as $f(x)$, but this does not seem to be right. Can you explain how to calculate the correct variance as given above, and how it is different from a linear combination of two random variables?

Any help will be greatly appreciated. Thank you.


1 Answer 1


Let $\mathcal T$ be the type of $X$ (i.e. normal or not, $\mathcal N,\mathcal N'$). Using the law of total variance, we can write: $$\operatorname{var}(X)=\mathbb E[\operatorname{var}(X|\mathcal T)] + \operatorname{var}(\mathbb E[X|\mathcal T])$$

For the first term, $$\mathbb E[\operatorname{var}(X|\mathcal T)]=P(\mathcal N)\operatorname{var}(X|\mathcal N)+P(\mathcal N')\operatorname{var}(X|\mathcal N')=(1-\delta)\sigma^2+\delta\tau^2$$

For the second term, $\mathbb E[X|\mathcal T]$ is a function of $\mathcal T$, and it has two options (i.e. either $\mu$ with probability $1-\delta$ or $\theta$ with probability $\delta$). You can directly calculate the variance using this, but this is a scaled and shifted version of a Bernoulli RV, say $Z$. So,

$$f(\mathcal T)=(\theta-\mu)Z+\mu$$

If $Y$ is $1$, $f(\mathcal T)$ is $\theta$, else $\mu$. And, let $Z$ be $1$ with probability $\delta$. $$\operatorname{var}((\theta-\mu)Z+\mu)=(\theta-\mu)^2\operatorname{var}(Z)=(\theta-\mu)^2\delta(1-\delta)$$

If you substitute all of these, you get the reported expression.

Regarding the linear combination, you can do that for the PDFs: $$f_X(x)=(1-\delta) f_{Y_1}(x)+\delta f_{Y_2}(x)$$ And, you could calculate the variance using this expression as well. But, it'd be more cumbersome. The correct representation of the RV $X$ in terms of $Y_i$ is the following:

$$X=\begin{cases}Y_1&\text{with prob. } &1-\delta\\Y_2 & \text{with prob. } &\delta\end{cases}$$

  • $\begingroup$ Thank you for your help! Found it very helpful. $\endgroup$
    – Siho Park
    Commented Dec 2, 2020 at 13:16

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