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])

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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)

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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<K\right) &=& Pr\left(\frac{S_t}{S_0} < \frac{K}{S_0}\right) \\ \\ \displaystyle &=& Pr\left(\ln{\frac{S_t}{S_0}} < \ln{\frac{K}{S_0}}\right) \\ \\ \displaystyle &=& Pr\left(Z< \frac{\ln{\frac{K}{S_0}} - \mu(t)}{\sigma(t)}\right) \\ \\ \displaystyle &=& Pr\left(Z<\frac{\ln{\frac{K}{S_0}} - \left(\alpha - \delta - \frac{1}{2}\sigma^2\right)t}{\sigma\sqrt{t}}\right) \end{array}

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_t<K\right] &=& S_0e^{(\alpha - \delta)t}\frac{\mathcal{N}(-d_1)}{\mathcal{N}(-d_2)} \\ \\ \displaystyle E\left[S_t|S_t>K\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

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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<X<d \\ 0 &\mbox{ otherwise} \end{array} \right.
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.

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Filed under Coverage Modifications, Deductibles, Limits, Probability, Severity Models

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}

 

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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}

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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} 

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