An insurer offers a travelers insurance policy. Losses under the policy are uniformly distributed on the interval $[0, 5]$.
The insurer reimburses a policyholder for a loss up to a maximum of $4$.
Determine the cumulative distribution function, $F$, of the benefit that the insurer pays a policyholder who experiences exactly one loss under the policy.
My attempt:
Let $Y$ represent the payment RV. Then $Y = x$ if $0\le x<4$ and $Y = 4$ if $4\le x<5$. So the cumulative distribution function is
$$F_Y(x) = \begin{cases}
\displaystyle\int_0^x t\; dt = \frac{x^2}{2} \;\;\text{if $0\le x<4$}\\
\displaystyle\int_4^x 4\; dt = 4(x-4) \;\;\text{if $4\le x<5$}
\end{cases}$$
This is not the correct answer. Can someone please suggest where I have gone wrong?
Best Answer
$$F_B(b) = \begin{cases} 0, & \text{if $b<0$ } \\ \frac{b}{5}, & \text{if $0\leq b<4$ } \\ 1, & \text{if $b\geq 4$ } \end{cases}$$
Observe that your CDF has "a jump" in $b=4$ thus the random varaible $B$ is not absolutely continuous and for this reason it does not admit a density
Uniform rv in $[0;5]$ has the following density $f_X(x)=\frac{1}{5}\cdot\mathbb{1}_{[0;5]}(x)$ thus the CDF can be calculated with integrals in the following way
$$F_B(b)=\frac{1}{5}\int_0^b dt=\frac{b}{5}$$
The second part of the CDF is
$$\frac{4}{5}+\frac{1}{5}\int_4^5 dt=1$$