problem 20

5.4.1 problem 20

(a)

\begin{matrix} \begin{matrix} Φ (z) & = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{z} e^{- x^{2} ∕ 2} dx = \frac{1}{\sqrt{2 π}} (\int_{- \infty}^{0} e^{- x^{2} ∕ 2} dx + \int_{0}^{z} e^{- x^{2} ∕ 2} dx) \end{matrix} \end{matrix}

We know from the Normal PDF that

\int_{- \infty}^{\infty} e^{- x^{2} ∕ 2} dx = \sqrt{2 π}

. Since the integrand is an even function, the integral from

- \infty

0

is half this value

Φ (z) = \frac{1}{2} + \frac{1}{\sqrt{2 π}} \int_{0}^{z} e^{- x^{2} ∕ 2} dx

By applying the variable substitution $u = x ∕ \sqrt{2}$ in the integral, we obtain $du = dx ∕ \sqrt{2}$ . The lower and upper limits of integration become $0$ and $z ∕ \sqrt{2}$ , respectively.

\begin{matrix} \begin{matrix} Φ (z) & = \frac{1}{2} + \frac{1}{\sqrt{2 π}} \int_{0}^{z ∕ \sqrt{2}} e^{- u^{2}} \sqrt{2} du = \frac{1}{2} + \frac{1}{2} \frac{2}{\sqrt{π}} \int_{0}^{z ∕ \sqrt{2}} e^{- u^{2}} du \end{matrix} \end{matrix}

Φ (z) = \frac{1}{2} + \frac{1}{2} erf (\frac{z}{\sqrt{2}})

(b)

erf (z) = \frac{2}{\sqrt{π}} \int_{0}^{z} e^{- x^{2}} dx

The integrand is an even function, so the integrals from 0 to $z$ and from $- z$ to 0 are the same

\begin{matrix} \begin{matrix} erf (z) & = \frac{2}{\sqrt{π}} \int_{- z}^{0} e^{- x^{2}} dx = - \frac{2}{\sqrt{π}} \int_{0}^{- z} e^{- x^{2}} dx \end{matrix} \end{matrix}

erf (- z) = - erf (z)

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