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Chapter 3: Problem 7

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

Expert verified
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

01

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: \[f(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}\].
02

Take the first derivative.

Differentiate the function w.r.t. \(x\) to obtain the first derivative. Apply the chain rule to get: \[f'(x)=-\frac{(x-\mu)}{\sigma^{3} \sqrt{2 \pi}} e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}\].
03

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: \[f''(x)=\frac{1}{\sigma^{3} \sqrt{2 \pi}}\left(1-\frac{(x-\mu)^{2}}{\sigma^{2}}\right)e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}\].
04

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 \(x=\mu+\sigma\).

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Most popular questions from this chapter

. Let \(X\) be \(N(0,1)\). Use the moment-generating-function technique to show that \(Y=X^{2}\) is \(\chi^{2}(1)\) Hint: \(\quad\) Evaluate the integral that represents \(E\left(e^{t X^{2}}\right)\) by writing \(w=x \sqrt{1-2 t}\), \(t<\frac{1}{2}\)

If \(M\left(t_{1}, t_{2}\right)\) is the mgf of a bivariate normal distribution, compute the covariance by using the formula $$ \frac{\partial^{2} M(0,0)}{\partial t_{1} \partial t_{2}}-\frac{\partial M(0,0)}{\partial t_{1}} \frac{\partial M(0,0)}{\partial t_{2}} $$ Now let \(\psi\left(t_{1}, t_{2}\right)=\log M\left(t_{1}, t_{2}\right) .\) Show that \(\partial^{2} \psi(0,0) / \partial t_{1} \partial t_{2}\) gives this covariance directly.

Let the \(\operatorname{pmf} p(x)\) be positive on and only on the nonnegative integers. Given that \(p(x)=(4 / x) p(x-1), x=1,2,3, \ldots\) Find \(p(x)\). Hint: \(\quad\) Note that \(p(1)=4 p(0), p(2)=\left(4^{2} / 2 !\right) p(0)\), and so on. That is, find each \(p(x)\) in terms of \(p(0)\) and then determine \(p(0)\) from $$ 1=p(0)+p(1)+p(2)+\cdots $$

Let \(F\) have an \(F\) -distribution with parameters \(r_{1}\) and \(r_{2} .\) Prove that \(1 / F\) has an \(F\) -distribution with parameters \(r_{2}\) and \(r_{1}\).

Show that the failure rate (hazard function) of the Pareto distribution is $$ \frac{h(x)}{1-H(x)}=\frac{\alpha}{\beta^{-1}+x} $$ Find the failure rate (hazard function) of the Burr distribution with cdf $$ G(y)=1-\left(\frac{1}{1+\beta y^{\tau}}\right)^{\alpha}, \quad 0 \leq y<\infty . $$ In each of these two failure rates, note what happens as the value of the variable increases.
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