Solutions - 5150-01

(1) Recall formula for expectation of a continuous random variable:

$$E[X]={\int }_{-\infty }^{\infty }x{f}_X(x)dx$$

---

(2) Use formula to find an equation relating $a$ and $b$:

$$\begin{array}{cc}{\int }_0^{1}x(a+b{x}^2)dx& =\frac{7}{10}\\ {[\frac{a{x}^2}{2}+\frac{b{x}^4}{4}]|}_0^1& =\frac{7}{10}\\ \frac{a}{2}+\frac{b}{4}=\frac{7}{10}& \end{array}$$

---

(3) Integrate the PDF to get a second equation:

Since integrating a PDF should yield $1$:

$$\begin{array}{cc}{\int }_0^{1}a+b{x}^{2}dx& =1\\ {[ax+\frac{b{x}^3}{3}]|}_0^1& =1\\ a+\frac{b}{3}& =1\end{array}$$

---

(4) Solve system of equations for $a$ and $b$:

Isolating $a$ in the second equation yields $a=1-\frac{b}{3}$.

Plugging this into the first equation yields $\frac{1-\frac{b}{3}}{2}+\frac{b}{4}=\frac{7}{10}$.

Solving for $b$ yields $b=2.4$, and thus $a=0.2$.

---

(5) Compute variance:

Using $\text{Var}[X]=E\left[X^2\right]-{\left(E[X]\right)}^2$, first compute $E\left[X^2\right]$:

$$\begin{array}{c}E\left[X^2\right]={\int }_0^1{x}^2(a+b{x}^2)dx\\ \\ \gg \gg {[\frac{a{x}^3}{3}+\frac{b{x}^5}{5}]|}_0^1\\ \\ \gg \gg \frac{a}{3}+\frac{b}{5}\gg \gg \frac{41}{75}\end{array}$$

So:

$$\text{Var}[X]\gg \gg \frac{41}{75}-{\left(\frac{7}{10}\right)}^2\gg \gg \frac{17}{300}$$