Question

Show that the constant \(c\) can be selected so that \(f(x)=c 2^{-x^{2}},-\infty

Short answer

The constant \(c\) that would satisfy the conditions for a normal pdf is \(\frac{1}{\sqrt{2 \pi}}\).

Step-by-step solution

Confirm Non-negativity

To show that \(f(x)\) is non-negative for all \(x\), observe that \(f(x) = c 2^{-x^{2}}\). Since \(c\) is a constant and \(2^{-x^{2}}\) is always positive for any real number \(x\), it follows that \(f(x) \geq 0\) for all \(x\). Thus, \(f(x)\) satisfies the first condition for a pdf.

Solve The Integral

To show that \(f(x)\) satisfies the second condition for a pdf, we need to prove that the integral of \(f(x)\) from \(-\infty\) to \(+\infty\) is equal to 1. Given \(f(x)=c 2^{-x^{2}}\), we can write this in terms of the exponential function using the provided hint. Notice that \(2=e^{\log 2}\), so \(f(x) = c e^{-x^{2} \log 2}\). The integral of \(f(x)\) from \(-\infty\) to \(+\infty\) is then equal to \(c\) times the integral of \(e^{-\frac{x^2}{2}}\) from \(-\infty\) to \(+\infty\). This integral is known to be equal to \(\sqrt{2 \pi}\). Therefore, we obtain \(c \sqrt{2 \pi}\).

Find The Constant \(c\)

In order for the integral of \(f(x)\) to be equal to 1 (thus satisfying the second condition for a pdf), we need to find the value of \(c\) such that \(c \sqrt{2 \pi}\) equals \(1\). Solving for \(c\), we find \(c = \frac{1}{\sqrt{2 \pi}}\). Thus, for this specific \(c\), \(f(x)\) satisfies the conditions for a normal pdf.