Question

Compute the hazard rate function of $X$when $X$is uniformly distributed over.$(0,a)$

Short answer

It's up to $\frac{1}{a-x}$if $x$in $(0,a)$otherwise its adequate iszero.

Step-by-step solution

Find the Variable function.

The hazard rate of the random variable is that the function

$\lambda \left(x\right)=\frac{f\left(x\right)}{1-F\left(x\right)}$

for every $X$specified$F\left(x\right)<1$otherwise it's capable of zero.

We know that the CDF of Uniform distribution over $(0,a)$

$F\left(x\right)=\frac{x}{a}$

Equation of hazard rate.

for$x\in (0,a)$left from that interval it's capable of zero, right from that interval is an adequate one.

Hence, we have got that

$$\begin{gathered} 1-F\left(x\right)=1 \\ -\frac{x}{a} \\ =\frac{a-x}{a} \end{gathered}$$

Finally, the hazard rate is adequate

$$\begin{gathered} \lambda \left(x\right)=\frac{\frac{1}{a}}{\frac{a-x}{a}} \\ =\frac{1}{a-x} \end{gathered}$$

For $x\in (0,a)$

Otherwise, it's up to zero.