Payment Distribution D8A3Dc

1. **Problem Statement:** Find the probability density function (pdf), cumulative distribution function (cdf), and expected value of the payment random variable $Y = (X - d)^+$, where $d$ is the deductible and $X$ is the loss random variable. 2. **General Setup:** The payment random variable is defined as: $$Y = \max(0, X - d) = (X - d)^+$$ The pdf and cdf of $Y$ depend on the distribution of $X$. --- ### Case 1: $X$ is exponential with mean $\theta$ 3. **Recall exponential distribution:** $$f_X(x) = \frac{1}{\theta} e^{-x/\theta}, \quad x \geq 0$$ $$F_X(x) = 1 - e^{-x/\theta}, \quad x \geq 0$$ 4. **Distribution of $Y$:** Since $Y = 0$ if $X \leq d$, and $Y = X - d$ if $X > d$, - Probability $Y=0$ is $P(X \leq d) = F_X(d) = 1 - e^{-d/\theta}$. - For $y > 0$, $P(Y \leq y) = P(X - d \leq y, X > d) = P(d < X \leq y + d) = F_X(y + d) - F_X(d)$. Thus, the cdf of $Y$ is: $$F_Y(y) = \begin{cases} 0, & y < 0 \\ 1 - e^{-d/\theta}, & y = 0 \\ \frac{F_X(y + d) - F_X(d)}{1 - F_X(d)} = 1 - e^{-y/\theta}, & y > 0 \end{cases}$$ 5. **Pdf of $Y$ for $y > 0$:** $$f_Y(y) = \frac{f_X(y + d)}{1 - F_X(d)} = \frac{\frac{1}{\theta} e^{-(y + d)/\theta}}{e^{-d/\theta}} = \frac{1}{\theta} e^{-y/\theta}$$ 6. **Expected value of $Y$:** $$E[Y] = E[(X - d)^+] = \int_d^\infty (x - d) f_X(x) dx$$ Calculate: $$E[Y] = \int_d^\infty (x - d) \frac{1}{\theta} e^{-x/\theta} dx$$ Use substitution $u = x - d$: $$E[Y] = \int_0^\infty u \frac{1}{\theta} e^{-(u + d)/\theta} du = e^{-d/\theta} \int_0^\infty u \frac{1}{\theta} e^{-u/\theta} du$$ The integral is the mean of an exponential with mean $\theta$: $$\int_0^\infty u \frac{1}{\theta} e^{-u/\theta} du = \theta$$ Therefore: $$E[Y] = e^{-d/\theta} \theta$$ --- ### Case 2: $X$ is uniform on $(0, \theta)$ with $d < \theta$ 7. **Recall uniform distribution:** $$f_X(x) = \frac{1}{\theta}, \quad 0 < x < \theta$$ $$F_X(x) = \frac{x}{\theta}, \quad 0 < x < \theta$$ 8. **Distribution of $Y$:** Again, $Y = 0$ if $X \leq d$, and $Y = X - d$ if $X > d$. - Probability $Y=0$ is $P(X \leq d) = F_X(d) = \frac{d}{\theta}$. - For $0 < y < \theta - d$: $$F_Y(y) = P(Y \leq y) = P(d < X \leq y + d) = \frac{y}{\theta - d}$$ 9. **Pdf of $Y$ for $0 < y < \theta - d$:** $$f_Y(y) = \frac{1}{\theta - d}$$ 10. **Expected value of $Y$:** $$E[Y] = \int_d^\theta (x - d) \frac{1}{\theta} dx = \frac{1}{\theta} \int_d^\theta (x - d) dx$$ Calculate the integral: $$\int_d^\theta (x - d) dx = \left[ \frac{(x - d)^2}{2} \right]_d^\theta = \frac{(\theta - d)^2}{2}$$ Therefore: $$E[Y] = \frac{(\theta - d)^2}{2 \theta}$$