Probability Density Function problem

$$f(x) = \begin{cases} 0 & \text{if x < 0} \\ x^2 & \text{if 0 ≤ x < \mu} \\ α + βx & \text{if \mu ≤ x ≤ 10} \\ 0 & \text{if x ≥ 10} \end{cases}$$

Considering $\Pr(x < µ) = 1/3$. Questions:

• Find $\mu$.

• Find $\alpha$.

• Find $\beta$.

• Find the expected value $E(x)$.

I was solving all questions related to probability density functions with not much difficulty, but this one is driving me nuts. What I have done so far:

$$\int_{0}^{\mu} x^2 \, dx + \int_{\mu}^{10} (\alpha + \beta x) \, dx = 1$$

which would lead to something like:

$$(\frac{x^3}{3} \bigg|_{0} ^{\mu}) + (\alpha \cdot x+ \beta \cdot \frac{x^2}{2}\bigg|_{\mu}^{10}) = 1$$

Anyone know how to proceed?

• If this was not a problem on probability at all, would $\int_0^\mu x^2\, dx$ equal $\frac{x^3}{3} \int_0^\mu dx$? – Dilip Sarwate Mar 18 '14 at 20:35
• It follows from $$\int_{0}^{\mu} x^2dx + \int_{\mu}^{10} (\alpha + \beta x)dx = 1$$ which would lead to something like: $$\frac{\mu^3}{3} + \alpha(10 -\mu) + \beta\frac{1}{2}\big( 100 - \mu^2 \big) = 1$$ while $\mathbb{P}\big( X < \mu \big) = 1/3$ gives us $$\int_{0}^{\mu} x^2dx = \frac{\mu^2}{3} = \frac{1}{3} \Rightarrow \mu = 1$$ So we could get a relation between $\alpha$ and $\beta$ $$\alpha + \frac{11}{2}\beta = \frac{2}{27} \,\, .$$ Are you sure that no other conditions is given ? – Chival Mar 18 '14 at 20:50
• Well, it only says that this function is a PDF (Probability Density Function).. nothing more.. This exercise was proposed in an economics master's degree course. – Argo Mar 18 '14 at 20:57
• @Chival "The area under the curve must equal $1$" – Dilip Sarwate Mar 18 '14 at 22:06
• @DilipSarwate, yes, but it is not enough for this question. – Chival Mar 19 '14 at 1:17