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The continuous random variable X, representing a certain component's lifetime in hours, is uniformly distributed between t = 10 hours and t = 20 hours.

a) What is its failure rate function λ(t), for 10 ≤ t ≤ 20?

b) What is its expected value?

So far I got the equation for $f(t) = \int^{\infty}_{-\infty} \frac{1}{\beta-\alpha} \mathop{d}t = F(t)$

$$f(t) = \int^{\infty}_{-\infty} \frac{1}{\beta-\alpha} \mathop{d}t = F(t)$$

*Pardon for my lack of knowledge in Latex.

[edit: the presentation is mostly okay, study the changes. NB: though the equation not quite right.]

1 Answers 1

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Close.   You have the right idea, but have not quite put it together.

You know that the probability density function for the uniform distributed random variable is $f(t)$. $$f(t) = \frac{1}{\beta-\alpha}\mathbf 1_{t\in(\alpha;\beta)}$$

From this you may obtain the cumulative distribution function, $F(t)$, if you do not already know what it is for a uniform distribution. (Always good to practice anyway.) $$\begin{align}F(t) ~ & = ~ \int_{-\infty}^t f(\tau)\mathop{\rm d} \tau \end{align}$$

$$\begin{align}F(t) ~ & = ~ \int_\alpha^t \frac{1}{\beta-\alpha}\mathop{\rm d} \tau\cdot \mathbf 1_{t\in[\alpha;\beta)} + \mathbf 1_{t\in[\beta;\infty)} \\[1ex] & = ~ \frac{t-\alpha}{\beta-\alpha}\mathbf 1_{t\in[\alpha;\beta)}+\mathbf 1_{t\in[\beta;\infty)} \\[1ex] & = ~\begin{cases}0&:& t< \alpha\\ (t-\alpha)/(\beta-\alpha) &:& \alpha\leq t\lt\beta \\ 1 &:& \beta\leq t\end{cases}\end{align}$$

The failure rate function is $\lambda(t)$ $$\lambda(t) = \frac{f(t)}{1-F(t)}$$

$\blacksquare$

PS: $\mathbf 1_{t\in A}$ is an indicator function, defined here as: $\mathbf 1_{t\in A} =\begin{cases}1 &:& t\in A\\ 0 &:& t\notin A\end{cases}$