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

There are two conditions that must be satisfied by $a$ and $b$: $${\text{Pr}}(-\infty <X<\infty )=\int _{a}^{\infty }e^{bx}dx=1$$ and $$\mathbb {E} (X)=\int _{a}^{\infty }xe^{bx}dx=0.$$ We begin by evaluating the first integral, and note right from the start that $b<0$, since otherwise the integral would be divergent. $$\lim _{t\rightarrow \infty }\int _{a}^{t}e^{bx}dx=\lim _{t\rightarrow \infty }{\frac {e^{bx}}{b}}{\Big |}_{a}^{t}=\lim _{t\rightarrow \infty }{\frac {e^{bt}}{b}}-{\frac {e^{ba}}{b}}=-{\frac {e^{ab}}{b}},$$ because $\lim _{t\rightarrow \infty }e^{bt}=0$ 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 $$\mathbb {E} (X)=\int _{a}^{\infty }xe^{bx}dx=\lim _{t\rightarrow \infty }\left({\frac {xe^{bx}}{b}}{\Big |}_{a}^{t}-\int _{a}^{t}{\frac {e^{bx}}{b}}dx\right)=\lim _{t\rightarrow \infty }\left({\frac {te^{bt}}{b}}-{\frac {ae^{ab}}{b}}-{\frac {e^{bt}}{b^{2}}}+{\frac {e^{ab}}{b^{2}}}\right).$$ But once again by the fact that $b<0$, $\lim _{t\rightarrow \infty }{\frac {te^{bt}}{b}}=0$ and $\lim _{t\rightarrow \infty }{\frac {e^{bt}}{b^{2}}}=0$, so $$\mathbb {E} (X)=-{\frac {ae^{ab}}{b}}+{\frac {e^{ab}}{b^{2}}}.$$ 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 $$a={\frac {1}{b}}=-{\frac {1}{e}}.$$