# Property of Random variable distributed as Standard normal with negative sign?

I'm newbie in probability. I'm studying the symmetry property of Standard Normal random variable. Example from the book "Introduction to probability" by Joseph K. Blitzstein and Jessica Hwang page 213 talks about "Symmetry of Z and -Z if Z ~ N(0,1) (Z has normal distribution with mean = 0 and Varience = 1), then -Z ~ N(0,1) as well."

"To see this, note that the CDF of -Z is:

$$P(-Z\le z) = P(Z\ge-z) = 1- \Phi(-z) = \Phi(z)$$

"

The equality above has 4 parts I do not understand the first two on the left which is this part "the CDF of -Z is $P(-Z\le z) = P(Z\ge-z)$"

How can the Z has negative version of it??????

I only study the CDF of Z for example, the definition is F(z) = CDF of Z then $F(z) = P(Z \le z)$

Which means when I say for example $F(3) = P(Z \le 3)$ = probability of Z less than or equal to 3, which is a number.

but when speaking about -Z then the CDF is $P(-Z\le z)$ Then for example if z = 3 the CDF of -Z is $F(3) = P(-Z \le 3)$ if we translate it to english then it is probability of -Z less than or equal to 3, which is none sense???? Can someone please tell me what it really means with consistent mathematical reason step by step explanation?

My second question is about the change from the most left statement to the second left statement which is $P(-Z\le z) = P(Z\ge-z)$ this seems like the author multiply the event inside by -1 why this is mathematically valid? Can we multiply any event by -1? Are we multiply event inside by -1 or we multiply the Random Variable by -1 which effect the event inside the perenthesis? which one is the correct mathematically consistant? Now it seems inconsistent, this is not what mathematic suppose to be.

• $Z \leq 3$ is an event, $Z$ is not an event, see it as a real random number. Nothing prevents you to multiply a real number with $-1$. If $-Z \leq 3$, almost surely you will have $Z \geq -3$ by using elementary mathematics. This characteristic is not proper to the normal distribution, you can define any random variable $U$ and multiply it by $-1$ to define a new variable. Commented Feb 23, 2017 at 14:46
• @Canardini From your answer, Assume I have 2 coin tosses. in one toss it can only be head or tail, so the sample space of 2 tosses is {HH, HT, TH, TT}. Now lets X be a random variable of number of Head from 2 tosses Ex. X(HH) = 2. So X has support 0,1,2 now I can use -X to manipulate sentence like "the number of head from 2 coin tosses minus the number of head from another 2 coin tosses." In this way, does the minus sign alter the support of X to be 0, -1, -2? or it doesn't alter the support of X and only after X crystalize into positive number, then we put minus sign in front of it? Commented Feb 24, 2017 at 12:52

## 2 Answers

It may help you to realize that a random variable is actually a function that maps a probability to a physical object (i.e a number in $\mathbb{R}$). So in this sense, it's completely consistent that you can multiply a random variable by -1. (@Canardini comment was also helpful)

Then

\begin{align*} \text{CDF of } -Z &= P(-Z \le z) \qquad \text{by definition of the CDF}\\ &= P(Z \ge -z) \qquad \text{by my comment above (also the comment by Canardini)}\\ &= 1 - \Phi(-z) \qquad \text{because the sum of all the probabilities is }1\\ &= \Phi(z) \end{align*}

• From your answer, Assume I have 2 coin tosses. in one toss it can only be head or tail, so the sample space of 2 tosses is {HH, HT, TH, TT}. Now lets X be a random variable of number of Head from 2 tosses Ex. X(HH) = 2. So X has support 0,1,2 now I can use -X to manipulate sentence like "the number of head from 2 coin tosses minus the number of head from another 2 coin tosses." In this way, does the minus sign alter the support of X to be 0, -1, -2? or it doesn't alter the support of X and only after X crystalize into positive number, then we put minus sign in front of it? Commented Feb 24, 2017 at 12:51

$Z$ refers to a random variable, so the notation $Z \leq z$ can be read as "the event that the rv Z is smaller than the constant $z$". For example in the Normal case, $Z$ is some random number in $\mathbb R$. There is nothing magical about taking the negative sign of that number.

Perhaps it is easier to see the result using densities. Let $f(x)$ be the Standard Normal density. $$P(Z \leq z) = \int^z_{-\infty}f(x)dx = \int^z_{-\infty}\frac{1}{\sqrt{2\pi}} e^{-x^2/2}dx\\ P(-Z \leq z) = \int^z_{-\infty}f(-x)dx = \int^z_{-\infty}\frac{1}{\sqrt{2\pi}} e^{-(-x)^2/2}dx = \int^z_{-\infty}\frac{1}{\sqrt{2\pi}} e^{-x^2/2}dx$$ For the second question, as was pointed out in the comments, $Z$ itself is not an event. However, $\lbrace Z \leq z \rbrace$ is an event, and you can do whatever algebra you want in order to modify the event. For example another event is $\lbrace Z +2 \leq z \rbrace$. Then you can ask: what is the probability of that event, i.e. $$P(Z +2 \leq z)$$

• From your answer, Assume I have 2 coin tosses. in one toss it can only be head or tail, so the sample space of 2 tosses is {HH, HT, TH, TT}. Now lets X be a random variable of number of Head from 2 tosses Ex. X(HH) = 2. So X has support 0,1,2 now I can use -X to manipulate sentence like "the number of head from 2 coin tosses minus the number of head from another 2 coin tosses." In this way, does the minus sign alter the support of X to be 0, -1, -2? or it doesn't alter the support of X and only after X crystalize into positive number, then we put minus sign in front of it? Commented Feb 24, 2017 at 12:56
• @user3270418 by setting $Y = -X$ you define a new random variable with support $\lbrace 0, -1, -2 \rbrace$. In general you are dealing with functions of random variables, i.e. $Y = f(X)$
– user126540
Commented Feb 24, 2017 at 13:05