Science:Math Exam Resources/Courses/MATH101 C/April 2025/Question 09/Solution 1

There are two conditions that must be satisfied by $a$ and $b$: $$\text{Pr}(-\mathrm{\infty }<X<\mathrm{\infty })=\underset{a}{\overset{\mathrm{\infty }}{\int }}{e}^{bx}dx=1$$ and $$\mathbb{𝔼}(X)=\underset{a}{\overset{\mathrm{\infty }}{\int }}x{e}^{bx}dx=0.$$ We begin by evaluating the first integral, and note right from the start that $b<0$, since otherwise the integral would be divergent. $$\underset{t\to \mathrm{\infty }}{\lim }\underset{a}{\overset{t}{\int }}{e}^{bx}dx=\underset{t\to \mathrm{\infty }}{\lim }\frac{e^{bx}}{b}{|}_a^t=\underset{t\to \mathrm{\infty }}{\lim }\frac{e^{bt}}{b}-\frac{e^{ba}}{b}=-\frac{e^{ab}}{b},$$ because $\underset{t\to \mathrm{\infty }}{\lim }{e}^{bt}=0$ since $b<0$. The condition that $f(x)$ is a probability density yields that $$-\frac{e^{ab}}{b}=1,\text{or}{e}^{ab}=-b.$$ Next, we use integration by parts to evaluate the expected value. We have $$\mathbb{𝔼}(X)=\underset{a}{\overset{\mathrm{\infty }}{\int }}x{e}^{bx}dx=\underset{t\to \mathrm{\infty }}{\lim }(\frac{x{e}^{bx}}{b}{|}_a^t-\underset{a}{\overset{t}{\int }}\frac{e^{bx}}{b}dx)=\underset{t\to \mathrm{\infty }}{\lim }(\frac{t{e}^{bt}}{b}-\frac{a{e}^{ab}}{b}-\frac{e^{bt}}{b^2}+\frac{e^{ab}}{b^2}).$$ But once again by the fact that $b<0$, $\underset{t\to \mathrm{\infty }}{\lim }\frac{t{e}^{bt}}{b}=0$ and $\underset{t\to \mathrm{\infty }}{\lim }\frac{e^{bt}}{b^2}=0$, so $$\mathbb{𝔼}(X)=-\frac{a{e}^{ab}}{b}+\frac{e^{ab}}{b^2}.$$ But remember that $e^{ab}=-b$, which allows us to simplify the above expression as $\mathbb{𝔼}(X)=a-\frac{1}{b}$. Using that $\mathbb{𝔼}(X)=0$, we obtain $$a-\frac{1}{b}=0,\text{or}ab=1.$$ Now, because $e^{ab}=-b$ and $ab=1$, we get $$b=-e^{ab}=-e^1=-e$$ and $$a=\frac{1}{b}=-\frac{1}{e}.$$