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Chapter 3: Problem 5

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

Short Answer

Expert verified
The constant \(c\) that makes \(f(x)=c2^{-x^{2}}\) satisfy the conditions of a normal pdf is \(c=\sqrt{\log 2/\pi}\).

Step by step solution

01

Convert 2 in terms of exponential function

Firstly, convert \(2\) as \(e^{\log 2}\). As a result, the function \(f(x)\) can be rewritten as \(f(x) = ce^{-x^{2} \log 2}\)
02

Confirming non-negativity

The function \(f(x) = ce^{-x^{2} \log 2}\) is a non-negative function since the exponential function is always positive and c is also chosen to be positive. This confirms the first condition of being a pdf.
03

Calculate integral

To satisfy the second condition, integrate the function from \(-\infty\) to \(+\infty\), which is equal to 1. So, calculate \(\int_{-\infty}^\infty ce^{-x^{2} \log 2} dx\). Simplify this by completing the square in the exponent and use the formula for a Gaussian integral, which is \(\sqrt{\pi/a}\) for \(\int_{-\infty}^\infty e^{-ax^{2}} dx\). Here, \(a=\log 2\), so the integral becomes \(c \sqrt{\pi/\log 2}\).
04

Find c

Setting the integral equal to 1 to find \(c\), we get \(c \sqrt{\pi/\log 2} = 1\). Solving for \(c\), we get \(c=\sqrt{\log 2/\pi}\).

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