Here is the distribution
$$
F(x) =
\begin{cases}
0 & \text{if } x<0 \\
0.3 & \text{if } 0 \leq x < 1 \\
\frac{2x+1}{10} & \text{if } 1 \leq x < 3 \\
1 & \text{if } x \geq 3
\end{cases}
$$
So, is the random variable X discrete or continuous?
Edit: I have never come across this kind of distributions before. Extensive googling also did not help much.
How do we calculate probability density function for this kind of distribution, will it contain discrete values as well as a continuous value?
Best Answer
These types of distributions are common in actuarial contexts, such as when we are considering an insurer's incurred loss on a policy that has a deductible. For instance, if ground-up losses (before deductible) on an automobile policy are exponentially distributed with a mean of \$1500, but the policy has a deductible of \$500, the insurer is only responsible for ground-up loss minus the deductible: if the insured experiences a loss that does not exceed the deductible--for instance, they get a small dent or a broken headlight--they don't file a claim because the cost of repair does not exceed the deductible, and the insurer's liability is zero. But if the policyholder has a major collision, the insured will file a claim and the insurer will pay the cost of repair minus the deductible.
So, we see that the insurer's liability is modeled by a mixed random variable $$Y = \begin{cases} 0, & X \le 500, \\ X - 500, & X > 500, \end{cases}$$ where $$f_X(x) = \frac{1}{1500}e^{-x/1500}, \quad x > 0$$ is the density for $X$. The probability distribution for $Y$ has a discrete portion since $$\Pr[Y = 0] = \Pr[X \le 500] = F_X(500) = 1 - e^{-1/3}.$$ The continuous portion of the distribution for $Y$ can be understood by noting that if $X > 500$, then $$F_Y(y) = \Pr[Y \le y] = \Pr[X - 500 < y] = \Pr[X \le y + 500] = F_X(y+500) = 1 - e^{-(y+500)/1500}.$$ Thus $$F_Y(y) = \begin{cases} 0, & y < 0 \\ 1 - e^{-(y+500)/1500}, & y \ge 0.\end{cases}$$ Note this function is not continuous at $y = 0$. Differentiation gives the continuous portion--i.e., the density--of the distribution for $Y$, but to specify things completely, we must write: $$\begin{align*} \Pr[Y = 0] &= 1 - e^{-1/3}, \\ f_Y(y) &= \frac{1}{1500}e^{-(y+500)/1500}, \quad y > 0. \end{align*}$$ I leave as an exercise for the reader to consider how we might compute the expected value and variance of $Y$, as well as how we might model a situation where the deductible is not ordinary as in this example, but a franchise deductible--i.e., the deductible is waived if a loss exceeding that amount occurs. It is also interesting to consider the case of policy limits--i.e., what happens to $Y$ if the insurer only pays up to a certain amount, say $L = 100000$.