problem 52

5.2.3 problem 52

a. From the definition of expectation

E [X] = \int_{- \infty}^{\infty} xf (x) dx = \int_{0}^{\infty} x^{2} e^{- x^{2} ∕ 2} dx

The integral in the above equation is similar to the integral that arises when calculating the second moment of a standard Normal distribution. Let $Z \sim N (0, 1)$ , with PDF $φ$ . We know that $E [Z] = 0$ and $Var (Z) = 1$ . From the definition of variance,

Var (Z) = E [Z^{2}] - {(E [Z])}^{2} = E [Z^{2}] - 0 = 1

E [Z^{2}] = \int_{- \infty}^{\infty} z^{2} φ (z) dz = \int_{- \infty}^{\infty} z^{2} \frac{1}{\sqrt{2 π}} e^{- z^{2} ∕ 2} dz = 1

\int_{- \infty}^{\infty} z^{2} e^{- z^{2} ∕ 2} dz = \sqrt{2 π}

Since the integrand is an even function, the integral in $(0, \infty)$ is half this value. Therefore,

E [X] = \frac{\sqrt{2 π}}{2} = \sqrt{\frac{π}{2}}

\begin{matrix} E [X^{2}] & = \int_{0}^{\infty} x^{2} f (x) dx = \int_{0}^{\infty} x^{3} e^{- \frac{x^{2}}{2}} dx = \int_{0}^{\infty} 2 u e^{- u} du \end{matrix}

The last integral is similar to the integral that arises when calculating the expected value of $Y \sim Expo (1)$ . We know that $E [Y] = 1$ ,

E [Y] = \int_{0}^{\infty} y e^{- y} dy = 1

It then follows that $E [X^{2}] = 2$

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