Question

Show that the graph of a pdf \(N\left(\mu, \sigma^{2}\right)\) has points of inflection at \(x=\mu-\sigma\) and \(x=\mu+\sigma\).

Short answer

The points of inflection for the graph of a pdf of a Normal Distribution are at \(x=\mu-\sigma\) and \(x=\mu+\sigma\).

Step-by-step solution

Write out the formula for the pdf of a Normal Distribution.

The pdf of a Normal Distribution \(N\left(\mu, \sigma^{2}\right)\) is given by the formula: \[f(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}\].

Take the first derivative.

Differentiate the function w.r.t. \(x\) to obtain the first derivative. Apply the chain rule to get: \[f'(x)=-\frac{(x-\mu)}{\sigma^{3} \sqrt{2 \pi}} e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}\].

Take the second derivative.

Now differentiate the first derivative \(f'(x)\) w.r.t. \(x\) to obtain the second derivative. Use the chain rule for differentiation. This gives: \[f''(x)=\frac{1}{\sigma^{3} \sqrt{2 \pi}}\left(1-\frac{(x-\mu)^{2}}{\sigma^{2}}\right)e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}\].

Find points of inflection.

The points of inflection are found by setting the second derivative equal to zero and solving for \(x\). This yields two solutions: \(x=\mu-\sigma\) and \(x=\mu+\sigma\). So the points of inflection are at \(x=\mu-\sigma\) and \(x=\mu+\sigma\).