Question

With $\mathrm{\Phi }\left(\mathrm{x}\right)$being the probability that a normal random variable with mean $0$and variance $1$is less than $\mathrm{x}$, which of the following are true:

(a)$\mathrm{\Phi }(-\mathrm{x})=\mathrm{\Phi }\left(\mathrm{x}\right)$

(b)$\mathrm{\Phi }\left(\mathrm{x}\right)+\mathrm{\Phi }(-\mathrm{x})=1$

(c)$\mathrm{\Phi }(-\mathrm{x})=1/\mathrm{\Phi }\left(\mathrm{x}\right)$

Short answer

a. $\mathrm{\Phi }(-\mathrm{x})=\mathrm{\Phi }\left(\mathrm{x}\right)$is false for probability that a normal random variable.

b.$\mathrm{\Phi }\left(\mathrm{x}\right)+\mathrm{\Phi }(-\mathrm{x})=1$is true for probability that a normal random variable.

c.$\mathrm{\Phi }(-\mathrm{x})=1/\mathrm{\Phi }\left(\mathrm{x}\right)$is false for probability that a normal random variable.

Step-by-step solution

Explanation (part a)

a.

The cumulative distribution function of a normal random variable with mean $0$and variance $1$is localid="1649443303171" $\mathrm{\varphi }(-\mathrm{x})$. (that is, the standard normal random variable).

Because $\mathrm{\varphi }\left(\mathrm{x}\right)$is not symmetric about $0$, localid="1649443289397" $\mathrm{\varphi }(-\mathrm{x})=\mathrm{\varphi }\left(\mathrm{x}\right)$is incorrect (as the probability density function of the normal random variable is symmetric about $0$instead of the cumulative distribution function).

The equation $\mathrm{\varphi }(-\mathrm{x})=1-\mathrm{\varphi }\left(\mathrm{x}\right)$is correct in general, while $\mathrm{\varphi }(-\mathrm{x})=\mathrm{\varphi }\left(\mathrm{x}\right)$is only correct when $\mathrm{x}=0$(which also corresponds to $\mathrm{\varphi }=0.5$).

Explanation (part b)

b.

$\mathrm{\varphi }\left(\mathrm{x}\right)+\mathrm{\varphi }(-\mathrm{x})=1$Because the equation $\mathrm{\varphi }(-\mathrm{x})=1-\mathrm{\varphi }\left(\mathrm{x}\right)$holds for the standard normal random variable, is correct.

We also notice that $\mathrm{\varphi }\left(\mathrm{x}\right)+\mathrm{\varphi }(-\mathrm{x})=1$resulted by adding $\mathrm{\varphi }\left(\mathrm{x}\right)$ in equation$\mathrm{\varphi }(-\mathrm{x})=1-\mathrm{\varphi }\left(\mathrm{x}\right)$

Explanation (part c)

c.

Because $\mathrm{\varphi }(-\mathrm{x})$and $\mathrm{\varphi }\left(\mathrm{x}\right)$both represent probabilities, the equation$\mathrm{\varphi }(-\mathrm{x})=\frac{1}{\mathrm{\varphi }\left(\mathrm{x}\right)}$is incorrect.

This means that $\mathrm{\varphi }(-\mathrm{x})$and $\mathrm{\varphi }\left(\mathrm{x}\right)$are both values between $0$and$1$ (inclusive).