Standard Deviation of normal distribution transformation - Wilcox book explanation

Question

I started reading the book of Rand Wilcox "Introduction to Robust Estimation and Hypothesis testing" (4th Edition).

In the first chapter of the book, it is written:

Let $X$ be a standard normal random variable having distribution $\Phi(x) = \Pr(X \le x).$ Then for any constant $K \gt 0, \Phi(x/K)$ is a normal distribution with standard deviation $K.$

Am I missing something or this statement is incorrect? The standard deviation will be $1/K$ and not $K,$ right?

I am asking because the example continues

Let $\epsilon$ be any constant, $0 \le \epsilon \le 1.$ The mixed normal distribution is $$H(x) = (1 − \epsilon)\Phi(x) + \epsilon\Phi(x/K),\tag{1.1}$$ which has mean $0$ and variance $1 − \epsilon + \epsilon K^2.$

I was wondering how he derives a s.d of $K$ when he divides the original standard normal distributed variable by $K$ (instead of multiplying).

Answer 1

This can be confusing, so let's look at a picture for guidance.

The left panel graphs the standard Normal distribution function $\Phi.$ (Because all distribution functions give, by definition, probabilities, we don't have to label the vertical axis: we know it extends from $0$ to $1.$) The vertical red line indicates where some arbitrary argument $x$ is.

The right panel shows how to think of the function $x\to \Phi(x/K)$ where, in this example, $K$ is approximately $3.$ Dividing by $K$ shrinks all points on the horizontal axis by a factor of $K,$ thereby moving $x$ two-thirds of the way towards the origin, $0,$ as shown.

As always with a graph, find the value $\Phi(x/K)$ by looking up from the point $x/K$ to the graph itself. This is the height that the graph of $x \to \Phi(x/K)$ must have at the point $x.$ That is,

to graph $x\to \Phi(x/K),$ we must find the height $\Phi(x/K)$ and pull it back to $x$ as shown by the red arrow.

The resulting graph of $x\to\Phi(x/K)$ is shown in black on the right hand panel.

I hope this makes it apparent that the effect is to expand the graph of $\Phi$ horizontally by a factor $K$ (relative to the origin, $0,$ of the horizontal axis). (When $K$ is negative, the effect is an expansion by a factor of $|K|$ followed by a flip around $0.$)

Of course, such an expansion multiplies the standard deviation by the same amount $|K|.$

---

Finally, https://stats.stackexchange.com/a/16609/919 shows that the variance of the mixture is the weighted mean of the two variances because the mixture components all have the same means (namely, $0$). Thus, the variance of a mixture of two equal-mean distributions $F$ and $G$ with weights $\epsilon$ and $1-\epsilon$ will be $\epsilon$ times the variance of $F$ plus $1-\epsilon$ times the variance of $G.$

In this application, $G=\Phi$ has unit variance and--as we have just seen--$F = x\to\Phi(x/K)$ has standard deviation $K,$ whence its variance is $K^2.$ Accordingly, the variance of the mixture is $\epsilon(K^2) + (1-\epsilon)(1^2),$ as claimed.