Assume that the probability for an angle $\phi$ is $P(\phi) = \lambda\phi^2$. The game pays $\$1000$ times the angle. What is $\lambda$?

Question

Wheel of fortune: Assume that the probability for an angle $\phi$ is $P(\phi) = \lambda\phi^2$. The game pays $\$1000$ times the angle.

a) What is $\lambda$?

b) Find the expectation and the variance of the game.

Comments: I've been thinking of this question for days. Other than an idea that there being $360$ degrees may be part of the solution, I don't know where to start.

I interpreted $\lambda$ as follows:

$P(φ) = \lambda\phi^2 = 1 \Rightarrow \lambda =\frac{1}{\phi^2} $

At this point answers are nice, however explanations will be crucial. I really want to know how a and b work, and if my ideas were in the right direction.

Answer 1

Hint: There must be $$ \int_0^{2\pi}\lambda\varphi^2\,d\varphi=\lambda\frac13(2\pi)^3=1, $$ hence $\lambda=\frac{3}{8\pi^3}$. If we have density, the rest are calculations.

Answer 2

You've almost got (1) down. What you want is to use $\mathsf P(2\pi)=1$ to determine the parameter, $\lambda$. (Note: this is assuming $\mathsf P$ is the cumulative probability function.)

For (2) you'll want to use

$\begin{align} \mathsf E(1000\varphi) &= \int_0^{2\pi} 1000\phi\mathsf P(\operatorname d \phi) & \text{where }\mathsf P(\operatorname d \phi)=\frac{\operatorname d \mathsf P(\phi)}{\operatorname d \phi}\operatorname d \phi \\ & = 1000\int_0^{2\pi} \phi \times(2\lambda\phi)\operatorname d \phi \\[2ex] \mathsf {Var}(1000\varphi) & = \left(\int_0^{2\pi} 10^6\phi^2\mathsf P(\operatorname d\phi) \right)-\mathsf E(1000\varphi)^2 \\ & =2\times10^6\left(\int_0^{2\pi} \lambda \phi^3\operatorname d \phi\right)-\mathsf E(1000\varphi)^2 \end{align}$