Ok, so here's the problem. Let the random variable X represent loss in 2005. The density function of X is exponential with mean equal to 1. Let the random variable Y represent loss in 2008. Y = 1.2X An insurance company covers the losses in 2008 up to a maximum payment of M. The insurance payment in Year 2008 is the random variable "W". We are told that in year 2005, the insurance company covered the full loss and that results in an expected payment of 1 (because the expectation of X is 1). We are told that in year 2008, because of the new maximum payment policy, the expectation for payment in 2008 (W) is equal to the expectation of payment in 2005. In other words, E[W]=1. We are asked to find M.
The solution in the book tells me that E[W] is equal to the survival function of Y integrated from 0 to M. My question, essentially, is why is this true? I understand that integrating the survival function gives us expectation, but in order to get the expectation of W, don't we need to integrate the survival function of W with respect to w? Why can we find the expectation of W by integrating the survival function of Y from 0 to M? I can solve the problem if I just accept this as true, because I can use the fact that Y = 1.2X to actually perform the integration based on the density function for X, but I need to understand why that is true. If it helps, the solution said to think of W = min(Y, M), which stands to reason, since W is either M (if Y>m) or Y (if Y<M). Any help with this would be greatly appreciated. I've been setting up integrals and rearranging equations for hours to try to understand this.
Note that $$W = \min(Y, M) = \begin{cases} Y, & Y \le M \\ M, & Y > M. \end{cases}$$ This is because if the loss in 2008 does not exceed the policy limit $M$, the insurer pays out the full loss on the claim, which is just $Y$. But if it does exceed the policy limit, then the payment is capped at the policy limit.
Next, we want to compute determine $M$ such that $$\operatorname{E}[W] = \operatorname{E}[X] = 1,$$ which is the condition that the expected payment in 2008 is equal to the expected payment in 2005. This is because there is no policy limit imposed on losses in 2005, and you already established that the expected payment is $1$.
Now we incorporate the relationship between $Y$ and $X$. Since $Y = 1.2X$, it follows that $$Y \sim \operatorname{Exponential}(\mu = 1.2),$$ that is to say, $$\Pr[Y \le y] = \Pr[1.2X \le y] = \Pr[X \le y/1.2] = 1 - e^{-y/1.2}, \quad y \ge 0.$$ So $$\operatorname{E}[W] = \operatorname{E}[\min(Y,M)] = \int_{y=0}^\infty \min(y,M) f_Y(y) \, dy.$$ When $0 \le y \le M$, we have $\min(y,M) = y$, and when $y > M$, we have $\min(y,M) = M$, therefore, $$\operatorname{E}[W] = \int_{y=0}^M y f_Y(y) \, dy + \int_{y=M}^\infty M f_Y(y) \, dy.$$ At this point we can substitute the relevant density and complete the computation, but this is not the essence of your question. Rather, you wish to understand why $$\operatorname{E}[W] = \int_{y=0}^M S_Y(y) \, dy,$$ To see this, observe that $$S_Y(y) = \Pr[Y > y] = \int_{t = y}^\infty f_Y(t) \, dt.$$ Hence $$\int_{y=0}^M S_Y(y) \, dy = \int_{y=0}^M \int_{t=y}^\infty f_Y(t) \, dt \, dy$$ by direct substitution. What is this region of integration? If we think of $(t,y)$ as being plotted in a Cartesian coordinate plane, with $t$ the horizontal axis and $y$ the vertical, then the region satisfying $(0 \le y \le M) \cap (y \le t \le \infty)$ is a horizontal "strip" of height $M$ in the first quadrant, with a triangular region removed. That is to say, this region is the intersection of the inequalities $y \ge 0$, $y \le M$, and $y \le t$. So if we are to interchange the order of integration, we must split up $t$ depending on whether $t \le M$ or $t > M$. In the first case, $t \le M$ gives us the integral $$\int_{t=0}^M \int_{y=0}^t f_Y(t) \, dy \, dt = \int_{t=0}^M f_Y(t) \int_{y=0}^t 1 \, dy \, dt = \int_{t=0}^M t f_Y(t) \, dt.$$ In the second case, $t > M$ gives $$\int_{t=M}^\infty \int_{y=0}^M f_Y(t) \, dy \, dt = \int_{t=M}^\infty f_Y(t) \int_{y=0}^M 1 \, dy \, dt = \int_{t=M}^\infty M f_Y(t) \, dt.$$ Adding these two pieces togther gies us $$\int_{y=0}^M S_Y(y) \, dy = \int_{t=0}^M t f_Y(t) \, dt + \int_{t=M}^\infty M f_Y(t) \, dt,$$ which, with an inconsequential change of the variable of integration from $t$ to $y$, demonstrates the desired equivalence with the first expression we derived for the expectation.