# Tag Archives: Expected Value

## Bonuses, Dividends, and Refunds

If a policy pays a reward to the participants when losses are below a certain level, this is a particular type of  problem which Weishaus calls a “bonus” problem.  The bonus, dividend, or refund amount is expressed as a maximum between 0 and the refunded amount.  For example, a 15% refund is paid on the difference between the $100 premium and the loss $L$. No refund is paid if losses exceed$100.  The refund amount $R$ can be expressed as

$R = 0.15 \max (0, 100-L)$

The key to finding the expected refund is knowing how to manipulate the max function and rewrite it as a min.  We can rewrite as

$\begin{array}{rll} R &=& 0.15 \max (100-100,100-L) \\ &=& 0.15(100-\min (100,L)) \end{array}$

So the expected value is given by

$E[R] = 0.15(100 - E[L \wedge 100])$

Filed under Coverage Modifications

## The Loss Elimination Ratio

If you impose a deductible $d$ on an insurance policy that you’ve written, what fraction of expected losses do you eliminate from your expected liability?  This is measured by the Loss Elimination Ratio $LER(d)$.

$\displaystyle LER(d) = \frac{E\left[X \wedge d\right]}{E\left[X\right]}$

Definitions:

1. Ordinary deductible $d$— The payment made by the writer of the policy is the loss $X$ minus the deductible $d$.  If the loss is less than $d$, then nothing is paid.
2. Franchise deductible $d_f$—  The payment made by the writer of the policy is the complete amount of the loss $X$ if $X$ is greater than $d_f$.
A common type of question considers what happens to LER if an inflation rate $r$ increases the amount of all losses, but the deductible remains unadjusted.  Let $X$ be the loss variable.  Then $Y=(1+r)X$ is the inflation adjusted loss variable.  If losses $Y$ are subject to deductible $d$, then
$\begin{array}{rll} \displaystyle LER_Y(d) &=& \frac{E\left[(1+r)X\wedge d\right]}{E\left[(1+r)X\right]} \\ \\ \displaystyle &=&\frac{(1+r)E\left[X\wedge \frac{d}{1+r}\right]}{(1+r)E\left[X\right]} \\ \\ &=& \frac{E\left[X \wedge \frac{d}{1+r}\right]}{E\left[X\right]}\end{array}$
Memorize:
$\displaystyle E\left[X \wedge d\right] = \int_0^d{x f(x) dx} + d\left(1-F(x)\right)$

## The Lognormal Distribution

Review: If $X$ is normal with mean $\mu$ and standard deviation $\sigma$, then

$Z = \displaystyle \frac{X-\mu}{\sigma}$

is the Standard Normal Distribution with mean 0 and standard deviation 1.  To find the probability $Pr(X \le x)$, you would convert $X$ to the standard normal distribution and look up the values in the standard normal table.

$\begin{array}{rll} Pr(X \le x) &=& Pr\left(\displaystyle \frac{X-\mu}{\sigma} \le \frac{x-\mu}{\sigma}\right) \\ \\ &=& \displaystyle Pr\left(Z \le \frac{x-\mu}{\sigma}\right) \\ \\ &=& \displaystyle \mathcal{N}\left(\frac{x-\mu}{\sigma}\right) \end{array}$

If $V$ is a weighted sum of $n$ normal random variables $X_i, i = 1, ..., n$, with means $\mu_i$, variance $\sigma^2_i$, and weights $w_i$, then

$\displaystyle E\left[\sum_{i=1}^n w_iX_i\right] = \sum_{i=1}^n w_i\mu_i$

and variance

$\displaystyle Var\left(\sum_{i=1}^n w_iX_i\right) = \sum_{i=1}^n \sum_{j=1}^n w_iw_j\sigma_{ij}$

where $\sigma_{ij}$ is the covariance between $X_i$ and $X_j$.  Note when $i=j$, $\sigma_{ij} = \sigma_i^2 = \sigma_j^2$.

Remember: A sum of random variables is not the same as a mixture distribution!  The expected value is the same, but the variance is not.  A sum of normal random variables is also normal.  So $V$ is normal with the above mean and variance.

Actuary Speak: This is called a stable distribution.  The sum of random variables from the same distribution family produces a random variable that is also from the same distribution family.

The fun stuff:
If $X$ is normal, then $Y = e^X$ is lognormal.  If $X$ has mean $\mu$ and standard deviation $\sigma$, then

$\begin{array}{rll} \displaystyle E\left[Y\right] &=& E\left[e^X\right] \\ \\ \displaystyle &=& e^{\mu + \frac{1}{2}\sigma^2} \\ \\ Var\left(e^X\right) &=& e^{2\mu + \sigma^2}\left(e^{\sigma^2} - 1\right)\end{array}$

Recall $FV = e^\delta$ where $FV$ is the future value of an investment growing at a continuously compounded rate of $\delta$ for one period.  If the rate of growth is a normal distributed random variable, then the future value is lognormal.  The Black-Scholes model for option prices assumes stocks appreciate at a continuously compounded rate that is normally distributed.

$S_t = S_0e^{R(0,t)}$

where $S_t$ is the stock price at time $t$, $S_0$ is the current price, and $R(0,t)$ is the random variable for the rate of return from time 0 to t.  Now consider the situation where $R(0,t)$ is the sum of iid normal random variables $R(0,h) + R(h,2h) + ... + R((n-1)h,t)$ each having mean $\mu_h$ and variance $\sigma_h^2$.  Then

$\begin{array}{rll} E\left[R(0,t)\right] &=& n\mu_h \\ Var\left(R(0,t)\right) &=& n\sigma_h^2 \end{array}$

If $h$ represents 1 year, this says that the expected return in 10 years is 10 times the one year return and the standard deviation is $\sqrt{10}$ times the annual standard deviation.  This allows us to formulate a function for the mean and standard deviation with respect to time.  Suppose we write

$\begin{array}{rll} \displaystyle \mu(t) &=& \left(\alpha - \delta -\frac{1}{2}\sigma^2\right)t \\ \sigma(t) &=& \sigma \sqrt{t} \end{array}$

where $\alpha$ is the growth factor and $\delta$ is the continuous rate of dividend payout.  Since all normal random variables are transformations of the standard normal, we can write $R(0,t) =\mu(t)+Z\sigma(t)$ . The model for the stock price becomes

$\displaystyle S_t = S_0e^{\left(\alpha - \delta - \frac{1}{2}\sigma^2\right)t + Z\sigma\sqrt{t}}$

In this model, the expected value of the stock price at time $t$ is

$E\left[S_t\right] = S_0e^{(\alpha - \delta)t}$

Actuary Speak: The standard deviation $\sigma$ of the return rate is called the volatility of the stock.  This term comes from expressing the rate of return as an Ito process. $\mu(t)$ is called the drift term and $\sigma(t)$ is called the volatility term.

Confidence intervals: To find the range of stock prices that corresponds to a particular confidence interval, we need only look at the confidence interval on the standard normal distribution then translate that interval into stock prices using the equation for $S_t$.

Example: For example $z=[-1.96, 1.96]$ represents the 95% confidence interval in the standard normal $\mathcal{N}(z)$.  Suppose $t = \frac{1}{3}$, $\alpha = 0.15$, $\delta = 0.01$, $\sigma = 0.3$, and $S_0 = 40$.  Then the 95% confidence interval for $S_t$ is

$\left[40e^{(0.15-0.01-\frac{1}{2}0.3^2)\frac{1}{3} + (-1.96)0.3\sqrt{\frac{1}{3}}},40e^{(0.15-0.01-\frac{1}{2}0.3^2)\frac{1}{3} + (1.96)0.3\sqrt{\frac{1}{3}}}\right]$

Which corresponds to the price interval of

$\left[29.40,57.98\right]$

Probabilities: Probability calculations on stock prices require a bit more mental gymnastics.

$\begin{array}{rll} \displaystyle Pr\left(S_t

Conditional Expected Value: Define

$\begin{array}{rll} \displaystyle d_1 &=& -\frac{\ln{\frac{K}{S_0}} - \left(\alpha - \delta + \frac{1}{2}\sigma^2\right)t}{\sigma\sqrt{t}} \\ \\ \displaystyle d_2 &=& -\frac{\ln{\frac{K}{S_0}}- \left(\alpha - \delta - \frac{1}{2}\sigma^2\right)t}{\sigma\sqrt{t}} \end{array}$

Then

$\begin{array}{rll} \displaystyle E\left[S_t|S_tK\right] &=& S_0e^{(\alpha - \delta)t}\frac{\mathcal{N}(d_1)}{\mathcal{N}(d_2)} \end{array}$

This gives the expected stock price at time $t$ given that it is less than $K$ or greater than $K$ respectively.

Black-Scholes formula: A call option $C_t$ on stock $S_t$ has value $\max\left(0,S_t - K\right)$ at time $t$.  The option pays out if $S_t > K$.  So the value of this option at time 0 is the probability that it pays out at time $t$, discounted by the risk free interest rate $r$, and multiplied by the expected value of $S_t - K$ given that $S_t > K$.  In other words,

$\begin{array}{rll} \displaystyle C_0 &=& e^{-rt}Pr\left(S_t>K\right)E\left[S_t-K|S_t>K\right] \\ \\ &=& e^{-rt}\mathcal{N}(d_2)\left(E\left[S_t|S_t>K\right] - E\left[K|S_t>K\right]\right) \\ \\ &=& e^{-rt}\mathcal{N}(d_2)\left(S_0e^{(\alpha - \delta)t}\frac{\mathcal{N}(d_1)}{\mathcal{N}(d_2)} - K\right) \end{array}$

Black-Scholes makes the additional assumption that all investors are risk neutral.  This means assets do not pay a risk premium for being more risky.  Long story short, $\alpha - r = 0$ so $\alpha = r$.  So in the Black-Scholes formula:

$\begin{array}{rll} \displaystyle d_1 &=& -\frac{\ln{\frac{K}{S_0}} - \left(r - \delta + \frac{1}{2}\sigma^2\right)t}{\sigma\sqrt{t}} \\ \\ \displaystyle d_2 &=& -\frac{\ln{\frac{K}{S_0}}- \left(r- \delta - \frac{1}{2}\sigma^2\right)t}{\sigma\sqrt{t}} \end{array}$

Continuing our derivation of $C_0$ but replacing $\alpha$ with $r$,

$\begin{array}{rll} \displaystyle C_0 &=& e^{-rt}\mathcal{N}(d_2)\left(S_0e^{(r - \delta)t}\frac{\mathcal{N}(d_1)}{\mathcal{N}(d_2)} - K\right) \\ \\ &=& S_0e^{-\delta t}\mathcal{N}(d_1) - Ke^{-rt}\mathcal{N}(d_2)\end{array}$

For a put option $P_0$ with payout $K-S_t$ for $K>S_t$ and 0 otherwise,

$P_0 = Ke^{-rt}\mathcal{N}(-d_2) - S_0e^{-\delta t}\mathcal{N}(-d_1)$

These are the famous Black-Scholes formulas for option pricing.  When derived on the back of a cocktail napkin, they are indispensable for impressing the ladies at your local bar.  :p

## Expected Values for Insurance

Before I begin, please note: I hated this chapter.  If there are any errors please let me know asap!

A deductible $d$ is an amount that is subtracted from an insurance claim.  If you have a $500 deductible on your car insurance, your insurance company will only pay damages incurred beyond$500.  We are interested in the following random variables: $(X - d)_+$ and $(X\wedge d)$.

Definitions:

1. Payment per Loss: $(X-d)_+ = \left\{ \begin{array}{ll} X-d &\mbox{ if } X>d \\ 0 &\mbox{ otherwise} \end{array} \right.$
2. Limited Payment per Loss:  $(X\wedge d) = \left\{ \begin{array}{ll} d &\mbox{ if } X>d \\ X &\mbox{ if } 0
Expected Values:
1. $\begin{array}{rll} E[(X-d)_+] &=& \displaystyle \int_{d}^{\infty}{(x-d)f(x)dx} \\ \\ &=& \displaystyle \int_{d}^{\infty}{S(x)dx} \end{array}$

2. $\begin{array}{rll} E[(X\wedge d)] &=& \displaystyle \int_{0}^{d}{xf(x)dx +dS(x)} \\ \\ &=& \displaystyle \int_{0}^{d}{S(x)dx} \end{array}$
We may also be interested in the payment per loss, given payment is incurred (payment per payment) $X-d|X>d$.
By definition:
$E[X-d|X>d] = \displaystyle \frac{E[(X-d)_+]}{P(X>d)}$
Since actuaries like to make things more complicated than they really are, we have special names for this expected value.  It is denoted by $e_X(d)$ and is called mean excess loss in P&C insurance and $\displaystyle {\mathop{e}\limits^{\circ}}_d$ is called mean residual life in life insurance.  Weishaus simplifies the notation by using the P&C notation without the random variable subscript.  I’ll use the same.
Memorize!
1. For an exponential distribution,
$e(d) = \theta$
2. For a Pareto distribution,
$e(d) = \displaystyle \frac{\theta +d}{\alpha - 1}$
3. For a single parameter Pareto distribution,
$e(d) = \displaystyle \frac{d}{\alpha - 1}$
Useful Relationships:
1. $\begin{array}{rll} E[X] &=& E[X\wedge d] + E[(X-d)_+] \\ &=& E[X\wedge d] + e(d)[1-F(d)] \end{array}$
Actuary Speak (important for problem comprehension):
1. The random variable $(X-d)_+$ is said to be shifted by $d$ and censored.
2. $e(d)$ is called mean excess loss or mean residual life.
3. The random variable $X\wedge d$ can be called limited expected value, payment per loss with claims limit, and amount not paid due to deductible.  $d$ can be called a claims limit or deductible depending on how it is used in the problem.
4. If data is given for $X$ with observed values and number of observations or probabilities, the data is called the empirical distribution.  Sometimes empirical distributions may be given for a problem, but you are still asked to assume an parametric distribution for $X$.

## The Bernoulli Shortcut

If $X$ has a Standard Bernoulli Distribution, then it can only have values 0 or 1 with probabilities $q$ and $1-q$.  Any random variables that can only have 2 values is a scaled and translated version of the standard bernoulli distribution.

Expected Value and Variance:

For a standard bernoulli distribution, $E[X] = q$ and $Var(X) = q(1-q)$.  If $Y$ is a random variable that can only have values $a$ and $b$ with probabilities $q$ and $(1-q)$ respectively, then

$\begin{array}{rl} Y &= (a-b)X +b \\ E[Y] &= (a-b)E[X] +b \\ Var(Y) &= (a-b)^2Var(X) \\ &= (a-b)^2q(1-q) \end{array}$

Filed under Probability

## Mixture Distributions

Finite:  A finite mixture distribution is described by the following cumulative distribution function:

$F(x) = \displaystyle \sum_{i=1}^n w_iF(x_i)$

Where $X$ is the mixture random variable, $X_i$ are the component random variables that make up the mixture, and $w_i$ is the weighting for each component.  The weights add to 1.

If $X$ is a mixture of 50% $X_1$ and 50% $X_2$, $F(x) = 0.5F(x_1) + 0.5F(x_2)$.  This is not the same as $X = 0.5X_1 +0.5X_2$.  The latter expression is a sum of random variables NOT a mixture!

Moments and Variance:

$\begin{array}{rl} E(X^t) &= \displaystyle \sum_{i=1}^n w_iE(X_i^t) \\ Var(X) &= E(X^2) - E(X)^2 \\ &= \displaystyle \sum_{i=1}^n w_iE(X^2) - \left(\sum_{i=1}^n w_iE(X)\right)^2 \end{array}$

Filed under Probability

## Variance and Expected Value Algebra

Linearity of Expected Value: Suppose $X$ and $Y$ are random variables and $a$ and $b$ are scalars.  The following relationships hold:

$E[aX+b] = aE[X]+b$

$E[aX+bY] = aE[X] +bE[Y]$

Variance:

$Var(aX+bY) = a^2Var(X)+2abCov(X,Y)+b^2Var(Y)$

Suppose $X_i$ for $i=\left\{1\ldots n\right\}$ are $n$ independent identically distributed (iid) random variables.  Then $Cov(X_i,X_j) = 0$ for $i\ne j$ and

$\displaystyle Var\left({\sum_{i=1}^n X_i}\right) = \sum_{i=1}^n Var(X_i)$

Example:

$X$ is the stock price of AAPL at market close.  $Y$ is the sum of closing AAPL stock prices for 5 days.  Then

$\begin{array}{rl} Var(Y) &= \displaystyle \sum_{i=1}^5 Var(X_i) \\ &= 5Var(X) \end{array}$.

Contrast this with the variance of $Z = 5X$.  In other words, $Z$ is a random variable that takes a value of 5 times the price of AAPL at the close of any given day.  Then

$\begin{array}{rl} Var(Z) &= Var(5X) \\ &=5^2Var(x) \end{array}$

The distinction between $Y$ and $Z$ is subtle but very important.

Variance of a Sample Mean:

In situations where the sample mean $\bar{X}$ is a random variable over $n$ iid observations (i.e. the average price of AAPL over 5 days), the following formula applies:

$\begin{array}{rl} Var(\bar{X}) &= \displaystyle Var\left(\frac{1}{n} \displaystyle \sum_{i=1}^n X_i\right) \\ &= \displaystyle \frac{nVar(X)}{n^2} \\ &= \displaystyle \frac{Var(X)}{n} \end{array}$