Expected value of random variable with piecewise function as its distribution

Question

Exercise. Claim sizes on an insurance policy have the following distribution:

$$F(x)=\begin{cases}0&x\le0\\0.0002x&0<x<1000\\0.4&x=1000\\1-0.6e^{(x-1000)/2000}&x>1000\end{cases}$$

Calculate the expected claim size.

A) 1500

B) 1700

C) 1900

D) 2100

E) 2300

Correct me if I'm wrong, but this is a mixed distribution because there is a jump (discontinuity) in $F(x) \space$ at $\space x=1000$.

I know one way of doing it would be to integrate the survival function. But why wouldn't taking the derivative and integrating $\space xf(x) \space$ in each interval plus the point mass at $1000$ work?

For example, when $\space 0<x<1000, \space$ then $\space f(x) = .0002,\space $ so integrating $\space .0002x \space$ from $0,1000$ would give $.0001(1000)^2 = 100$.

The point mass I believe should have a probability of $.2$, since that's the height of the jump in $\space F(x) \space$ at $\space x=1000,\space$ so $.2(1000)$ to take into account the point mass.

Then do the same for the last piece, though that might be hard to integrate $\space F'(x)\cdot x. \space$ But I ran though it with technology and it's nowhere close to the answer it should be for the overall expected value.

Answer 1

Using your method, I end up with $2100$, so (D).

I have to assume your error is somewhere in your last step that you didn't include.

---

An alternative method is to write your random variable $X$ as a sum of more basic ones: $$X = 0.2 \cdot U_{[0,1000]} + 0.2\cdot\delta_{1000} + 0.6\cdot (R_{1000}E_{2000})$$ where $U_{[a,b]}$ is uniform on $[a,b]$, $\delta_{c}$ is a point-mass at $x=c$, $E_\lambda$ is an exponential with mean $\lambda$, and $R_x$ right-shifts the distribution by $x$.

Then $$\begin{align*}\mathbb{E}[X] &= 0.2\mathbb{E}[U_{[0,1000]}] + 0.2\mathbb{E}[\delta_{1000}] + 0.6\left(\mathbb{E}[E_{2000}] + 1000\right) \\ &= 0.2(500) + 0.2(1000) + 0.6(2000+1000) \\ &= 2100 \end{align*}$$

Answer 2

Doing it your way:
$$x f_X(x) = \begin{cases} 0, & x < 0 \\ 0.0002x, & 0 \le x < 1000 \\ 0.0003x e^{-(x-1000)/2000}, & x > 1000 \\ \end{cases}$$ with the added provision $\Pr[X = 1000] = \Pr[X \le 1000] - \Pr[X < 1000] = 0.4 - 0.2 = 0.2$. Therefore, $$\begin{align*} \operatorname{E}[X] &= \int_{x=0}^{1000} 0.0002 x \, dx + 1000\Pr[X = 1000] + \int_{x=1000}^\infty 0.0003x e^{-(x-1000)/2000} \, dx \\ &= 100 + 200 + 1800 \\ &= 2100. \end{align*}$$ The calculation of the last integral is accomplished via the substitution $$u = \frac{x-1000}{2000}, \quad x = 2000u + 1000, \quad dx = 2000 \, du, \tag{$*$}$$ hence $$\begin{align*} \int_{x=1000}^\infty 0.0003 xe^{-(x-1000)/2000} \, dx &= 600 \int_{u=0}^\infty (2u + 1)e^{-u} \, du \\&= 1200 \int_{u=0}^\infty u e^{-u} \, du + 600 \int_{u=0}^\infty e^{-u} \, du. \end{align*} $$ The first integral is $1$ since it is the integral of an exponential density with mean $1$. The second integral is also $1$ from simply performing the calculation. Therefore, the integral evaluates to $1800$ as claimed and the expectation is $2100$.

---

Doing it the survival curve way, we perform the same substitution $(*)$ as we did above: $$\begin{align*} \operatorname{E}[X] &= \int_{x=0}^\infty S_X(x) \, dx \\ &= \int_{x=0}^{1000} 1 - 0.0002x \, dx + \int_{x=1000}^\infty 0.6 e^{-(x-1000)/2000} \, dx \\ &= \left[x - 0.0001x^2\right]_{x=0}^{1000} + \int_{u=0}^\infty 1200 e^{-u} \, du \\ &= 1000 - 100 + 1200 \\ &= 2100. \end{align*}$$ I leave it to the reader to decide which approach is simpler.