Question

Let the random variable \(X\) have the pdf $$ f(x)=\frac{2}{\sqrt{2 \pi}} e^{-x^{2} / 2}, \quad 0

Short answer

The mean of \(X\) is \(1\), the variance of \(X\) is \(2\), the cumulative distribution function of \(X\) is \(1 - e^{-x^2/2}\), and the hazard function of \(X\) is \(\frac{2}{\sqrt{2\pi}}\).

Step-by-step solution

- Compute the mean

Start by finding the expected value \(E(X)\) also known as the mean, which is given by the integral \(\int_{0}^{\infty} x f(x) dx\). Substituting the given probability density function \(f(x)=\frac{2}{\sqrt{2 \pi}} e^{-x^{2} / 2}\), perform the necessary calculation. This involves integrating by parts and noting that \(\int_{0}^{\infty} e^{-x^2} dx = \sqrt{\pi/2}\), which then leads to the result \(E(X)=1\).

- Compute the variance

Next, find the variance of \(X\), which is defined as \(E(X^2) - (E(X))^2\). To do this, first calculate \(E(X^2)\) by performing the integral \(\int_{0}^{\infty} x^2 f(x) dx\). After the calculation, you will get that \(E(X^2)=3\). Now, subtract the square of the mean \((E(X))^2 = 1^2 = 1\) from \(E(X^2)\) to find the variance \(Var(X) = E(X^2) - (E(X))^2 = 3 - 1 = 2\).

- Compute the cumulative distribution function

The cumulative distribution function (CDF) is given by the integral of the density function. Thus, compute the integral \(\int_{0}^{x} f(s) ds\) in order to find the CDF. Use the definition of the error function \(erf(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt\), to simplify the integral and then obtain the cumulative distribution function \(F(x) = 1 - e^{-x^2/2}\).

- Compute the hazard function

Finally, compute the hazard function. The hazard function is \(\frac{f(x)}{1-F(x)}\), where \(f(x)\) is the probability density function and \(F(x)\) is the cumulative distribution function calculated in previous steps. Substituting the known functions and simplify to get hazard function \(h(x) = \frac{2}{\sqrt{2\pi}}\frac{e^{-x^2/2}}{e^{-x^2/2}} = \frac{2}{\sqrt{2\pi}}\).