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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).

whuber
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Arg
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    Everything you quote is incorrect. Even the definition in the second quotation is wrong! But since there seem to be typographical errors in the equations, I wonder whether these quotations correctly reflect what's actually in the book. Indeed, I have been able to see these passages on the Amazon site and in the original they are fully correct: you just haven't quoted them accurately! It is invalid to drop all Greek symbols such as "$\Phi:$" the result is either wrong or nonsensical in every case. – whuber Mar 22 '22 at 14:56
  • It's true, I missed the "Φ" - actually an error from the copy/paste as it was not recognizing the character and deleted it by mistake , but the rest is exactly as I quoted. See the Amazon site preview, in pages 2-3 (of the book) : https://www.amazon.com/Introduction-Estimation-Hypothesis-Statistical-Modeling/dp/012804733X?asin=012804733X&revisionId=&format=4&depth=1 .

    If we include the Φ ( Φ(x) = P(X ≤ x) , K > 0, Φ(x/K) , H(x) = (1 − ε)Φ(x) + εΦ(x/K) ), does the rest make sense? The standard deviation is K or 1/K for the Φ(x/K) distribution?

     – Arg Mar 24 '22 at 08:49
  • I have edited your post to reflect what's in your comment. – whuber Mar 24 '22 at 14:17

1 Answers1

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

Figure showing how the graph of the CDF changes

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.

whuber
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