Transformation of Cumulative Distribution Function Involving Deductibles The question is:

The cumulative distribution function for health care costs experienced by a policyholder is modeled by the function
\begin{align*}
    F(x)=\left\{\begin{matrix}
1-\mathit{e}^{-\frac{x}{100}} &  &x>0 \\ 
 0&  &\text{otherwise} 
\end{matrix}\right..
\end{align*}
The policy has a deductible of 20. An insurer reimburses the policyholder for 100$\%$ of health are costs between 20 and 120 less the deductible. Health care costs above 120 are reimbursed at 50$\%$.
Let $G$ be the cumulative distribution function of reimbursements given that the reimbursement is positive. Calculate $G(115)$.

Hello, I am trying to understand a solution I found online that gave

$G(y)=\left\{\begin{matrix}
0 &x<20 \\ 
x-20 &20\leq x\leq 120 \\ 
0.5(x-120) & x>120
\end{matrix}\right.$

where Y is the reimbursement and X is the cost.
I understand everything up to the third line of G(y). Why is it subtracting 120 from x?
 A: To solve the problem (I guess it's an SOA (P) question) efficiently, you don't need to spell out $G(y)$ for all $y > 0$. Instead, by law of total probability, you can solve $G(115)$ directly as follows:
\begin{align}
 & G(115) = P(Y \leq 115) \\
=& P(Y \leq 115 | X \leq 20)P(X \leq 20) + \\
 & P(Y \leq 115 | 20 < X \leq 120)P(20 < X \leq 120) + \\
 & P(Y \leq 115 | X > 120) P(X > 120) \\
=& P(X \leq 20) + \\
 & P(X - 20 \leq 115|20 < X \leq 120)P(20 < X \leq 120) + \\
 & P(100 + 0.5(X - 120) \leq 115 | X > 120)P(X > 120) \\
=& P(X \leq 20) + P(X \leq 135 | 20 < X \leq 120)P(20 < X \leq 120) +
P(X \leq 150 | X > 120)P(X > 120) \\
=& P(X \leq 20) + P(20 < X \leq 120) + P(120 < X \leq 150) \\
=& P(X \leq 150) = 1 -e^{-150/100} = 1 - e^{-1.5} = 0.7769.  
\end{align}
A: The description of $G$ as "the cumulative distribution function of reimbursements, given the reimbursement is positive" does not match the way it is defined in the subsequent quote.  Moreover, the LHS uses $G(y)$ but the RHS is only a function of $x$.  This makes no sense.  A cumulative distribution function must have a range in the interval $[0,1]$.
Instead, what we should do is define the claim payment random variable $Y$ in terms of the cost variable $X$.  That is to say, $Y$ is the amount the insurer will pay after applying the deductible and coinsurance limits for a given procedure cost $X$.
Clearly, $Y = 0$ if $X \le 20$.  Then, if $X > 20$, the amount paid is $100\%$ up to a cost of $X = 120$, so $Y = X - 20$ for $20 < X \le 120$, because the insurer first subtracts the deductible from the cost.
Then, if $X > 120$, the amount paid is $50\%$ of the excess above $120$; i.e., $Y = 100 + (X - 120)/2$.  So in summary, we have
$$Y = \begin{cases}0, & 0 \le X \le 20 \\
X - 20, & 20 < X \le 120 \\
\frac{X + 80}{2}, & 120 < X < \infty.
 \end{cases}$$
The function $G$ that you wrote makes no sense, because for instance, if $X = 121$, it would stipulate that $G = 0.5$, whereas if $X = 120$, then $G = 100$.  The insurer's reimbursement (claim payment) must be a nondecreasing function of the cost; it makes no sense that if the cost were slightly greater than $120$, then the insurer pays almost nothing.  As you can see from how I have constructed $Y$, the insurer's reimbursement is a continuous, nondecreasing function of $X$.
Now, what we need to do is compute $$G(115) = \Pr[Y \le 115].$$  This is what is literally meant by "the cumulative distribution function of reimbursements"--it is the probability that the insurer will pay at most $115$.  The easiest way to answer this question is to first observe that if $X \le 120$, then $Y \le 100$; in other words, if the cost is $120$ or less, the insurer will pay at most $100$, which satisfies the condition $Y \le 115$.  So what we need to do is find $X$ such that $115 = \frac{X+80}{2}$, or $X = 150$.  Therefore, $$G(115) = \Pr[Y \le 115] = \Pr[X \le 150] = 1 - e^{-150/100} \approx 0.77687.$$
