You design an insurance policy that pays a random amount Payment $= 1000 \cdot A$, where $A$ denotes the age at death. $A$ is assumed to have a continuous uniform distribution on $[50,110]$ (that is, a constant density function). Modify this policy to pay an amount Modified payment = $\begin{cases} 1000 \cdot A\; \text{if} \;A<90\\ 100,000 \;\text{if}\; A \geq 90 \end{cases}$

a) Find $E[X]$ and $\sigma$ of the original payment.

b) Find $E[X]$ and $\sigma$ of the modified payment.

I know the for the original payment the function jumps from 0-50 (like the step function) but I don't know how to compute $E[X]$ because my thinking for the 1st one was to just do $\dfrac{50}{110}(1000)\cdot 25.5 + \dfrac{60}{110}\cdot 1000 \displaystyle\int_{50}^{110} x$ which will give me my $E[X]$ for part a but that didn't work. So need some help understanding these sorts of problems.


An unreasonable distribution for the time of death!

We do the modified version, since it is a little more complicated. We let $T$be the age at time of death. By assumption the density function $f(t)$ of $T$ is $\frac{1}{60}$ on the interval $[50,110]$ and $0$ elsewhere.

The payment $X$ is $1000T$ if $50\le T\lt 90$, and $100000$ if $90\le T\le 110$. Let $g(t)=1000t$ if $50\le T\lt 90$, and let $g(t)=100000$ if $90\le t\le 110$. then $X=g(T)$, and therefore $$E(X)=\int_{50}^{110} g(t)f(t)\,dt.$$ Because of the shape of $g(t)$, we break up the integral into two parts, and get $$E(X)=\int_{50}^{90} \frac{1000 t}{60}\,dt +\int_{90}^{110} \frac{100000}{60}\,dt.$$ The integrations are very easy.

To find the standard deviation $\sigma$, we first find the variance $\sigma^2$ of $X$, and then take the square root. Recall that $\text{Var}(X)=E(X^2)-(E(X))^2$. We already know $E(X)$, so only need $E(X^2)$. This is $$\int_{50}^{110} (g(t))^2f(t)\,dt.$$ Just like before, we need to break up the integral, and we obtain $$E(X^2)=\int_{50}^{90} \frac{(1000 t)^2}{60}\,dt +\int_{90}^{110} \frac{(100000)^2}{60}\,dt.$$

The first problem is done in much the same way, except that there is no need to break up the integral. For $E(X)$, we integrate $\frac{1000 t}{60}$ from $50$ to $110$, and go through a similar procedure for $E(X^2)$.

  • $\begingroup$ For the 2nd part, why didn't you account for 0-50? I'm not saying its wrong buy why disregard it? $\endgroup$ – adam Dec 23 '13 at 5:43
  • $\begingroup$ The distribution given to us about the age of death says nobody dies before age $50$. Not exactly true! But we need to solve the problem under the unrealistic assumptions we are given. $\endgroup$ – André Nicolas Dec 23 '13 at 5:52
  • $\begingroup$ Oh ok I see because I thought we needed to do the entire ages from 0-50 as well. I guess that is what threw me off. So you split the integrals because the function is not differentiable at t=90 because of the jump right? $\endgroup$ – adam Dec 23 '13 at 5:58
  • $\begingroup$ It is not the non-differentiability, it is because the function $g(t)$ is defined by different formulas in the part of the world between $50$ and $90$ than in the part of the world between $90$ and $110$. Note that we need not worry about $\lt 50$ and $\gt 110$, since the density function is $0$ there. In principle we are integrating from $-\infty$ to $\infty$, but the density function in effect "lives" between $50$ and $110$. $\endgroup$ – André Nicolas Dec 23 '13 at 6:03

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