CODEWITHSUNDEEP
rstatisticsprobabilitynormal-distribution
Student_718
Student_718

Reputation: 1

Is pnorm(q = (x - $\mu$)/$\sigma$) ever different from pnorm(q = x, mean = $\mu$, sd = $\sigma$)?

It is the case that the probability density for a standardized and unstandardized random variable will differ. E.g., in R

dnorm(x = 0, mean = 1, sd = 2)
dnorm(x = (0 - 1)/2)

However, pnorm(q = 0, mean = 1, sd = 2) pnorm(q = (0 - 1)/2) yields the same value.

Are there any situations in which the Normal cumulative density function will yield a different probability for the same random variable when it is standardized versus unstandardized? If yes, is there a particular example in which this difference arises? If not, is there a general proof of this property?

Thanks so much for any help and/or insight!

Upvotes: 0

Views: 270

Answers (1)

DanY
DanY

Reputation: 6073

This isn't really a coding question, but I'll answer it anyway.

Short answer: yes, they may differ.

Long answer:

A normal distribution is usually thought of as y=f(x), that is, a curve over the domain of x. When you standardize, you are converting from units of x to units of z. For example, if x~N(15,5^2), then a value of 10 is 5 x-units less than the mean. Notice that this is also 1 standard deviation less than the mean. When you standardize, you convert x to z~N(0,1^2). Now, that example value of 10, when standarized into z-units, becomes a value of -1 (i.e., it's still one standard deviation less than the mean).

As a result, the area under the curve to the left of x=10 is the same as the area under the curve to the left of z=-1. In words, the cumulative probability up to those cut-offs is the same.

However, the height of curves is different. Let the normal distribution curves be f(x) and g(z). Then f(10) != g(-1). In code:

dnorm(10, 15, 5) != dnorm(-1, 0, 1)

The reason is that the act of standardizing either "spreads" or "squishes" the f(x) curve to make it "fit" over the new z domain as g(z).

Here are two links that let you visualize the spreading/squishing:

Hope this helps!

Upvotes: 1

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