lecture 18 continuous random variables

1

Suppose that the random variable $X$ has the probability density function

$$f(x)=\{\begin{array}{ll}m\cos (3x),& \frac{\pi }{6}<x<\frac{\pi }{3},\\ 0,& x\le \frac{\pi }{6}\text{or}x\ge \frac{\pi }{3}.\end{array}$$

Find the value of $m$ that makes $f$ a probability density function.

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Solution

A probability density function must satisfy

$${\int }_{-\infty }^{\infty }f(x)dx=1.$$

Because $f(x)=0$ outside $(\pi /6,\pi /3)$, we need

$${\int }_{\pi /6}^{\pi /3}m\cos (3x)dx=1.$$

Evaluate the integral:

$$m{\int }_{\pi /6}^{\pi /3}\cos (3x)dx=m{\left[\frac{1}{3}\sin (3x)\right]}_{\pi /6}^{\pi /3}=m\frac{1}{3}[\sin (\pi )-\sin (\frac{\pi }{2})]=m\frac{1}{3}(0-1)=-\frac{m}{3}.$$

Setting this equal to 1 gives

$$-\frac{m}{3}=1\Rightarrow m=-3.$$

Note that $\cos (3x)\le 0$ on $(\pi /6,\pi /3)$; with $m=-3$, the product $f(x)=m\cos (3x)$ is non-negative, as required.

$$\boxed{m=-3}$$

2

Problem (in LaTeX)

The continuous random variable $X$ has probability density function

$$f(x)=\{\begin{array}{ll}-\sin x+\frac{3}{2},& a\le x\le b,\\ 0,& \text{elsewhere}.\end{array}$$

Find the probability

$$\mathrm{Pr}(\frac{\pi }{6}<X<\frac{\pi }{3}).$$

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Because $f(x)=0$ outside $[a,b]$, the required probability is

$$\mathrm{Pr}(\frac{\pi }{6}<X<\frac{\pi }{3})={\int }_{\max \left(\frac{\pi }{6},a\right)}^{\min \left(\frac{\pi }{3},b\right)}[-\sin x+\frac{3}{2}]dx.$$

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When the whole interval $(\frac{\pi }{6},\frac{\pi }{3})$ lies inside $[a,b]$

If $a\le \frac{\pi }{6}\text{and}\frac{\pi }{3}\le b,$ we have

$$\begin{array}{rl}\mathrm{Pr}(\frac{\pi }{6}<X<\frac{\pi }{3})& ={\int }_{\pi /6}^{\pi /3}(-\sin x+\frac{3}{2})dx\\ & =[\cos x+\frac{3x}{2}{]}_{\pi /6}^{\pi /3}\\ & =(\cos \frac{\pi }{3}+\frac{3\pi }{6})-(\cos \frac{\pi }{6}+\frac{3\pi }{12})\\ & =\frac{1}{2}+\frac{\pi }{2}-\frac{\sqrt{3}}{2}-\frac{\pi }{4}\\ & =\frac{\pi }{4}+\frac{1-\sqrt{3}}{2}\approx \mathrm{0.419.}\end{array}$$

Hence, provided $\frac{\pi }{6}$ and $\frac{\pi }{3}$ both lie within the support $[a,b]$, the probability is

$$\boxed{\mathrm{Pr}(\frac{\pi }{6}<X<\frac{\pi }{3})=\frac{\pi }{4}+\frac{1-\sqrt{3}}{2}\approx 0.419}$$

Otherwise, take the integral only over the overlapping portion (or 0 if there is none).

3

Suppose that the random variable (X) has the probability density function.

Find the value of (a), correct to (2) decimal places, that makes (f) a probability density function.

$$f(x)=\{\begin{array}{ll}(x-1{)}^2+2,& a\le x\le 4,\\ 0,& x<a\text{or}x>4.\end{array}$$

Solution

For $f$ to be a valid probability-density function, its integral over the real line must be $1$.
Because $f(x)=0$ outside $[a,4]$,

$${\int }_{-\infty }^{\infty }f(x)dx={\int }_a^4((x-1{)}^2+2)dx=1.$$

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1. Evaluate the integral

$$\int ((x-1{)}^2+2)dx=\int (x^2-2x+1+2)dx=\int (x^2-2x+3)dx=\frac{x^3}{3}-x^2+3x+C.$$

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2. Apply the limits

$${\left[\frac{x^3}{3}-x^2+3x\right]}_a^4=\left(\frac{4^3}{3}-4^2+3\cdot 4\right)-\left(\frac{a^3}{3}-a^2+3a\right)=\frac{52}{3}-(\frac{a^3}{3}-a^2+3a).$$

Set this equal to $1$:

$$\frac{52}{3}-\left(\frac{a^3}{3}-a^2+3a\right)=1.$$

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3. Form the cubic equation

Multiply by $3$:

$$52-(a^3-3a^2+9a)=3⟹a^3-3a^2+9a-49=0.$$

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4. Solve for $a$

Testing integer endpoints shows the root lies between $3$ and $4$.
Using either a CAS, numerical solver or a few iterations of Newton’s method gives

$$a\approx \mathrm{3.91.}$$

(To two decimal places.)

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$$\boxed{a=3.91}$$