Posts tagged with lognormal

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November 9, 2009

Simulating the maximum…

Studying the distributional properties of the maximum is gaining more of attention nowadays given the relative failure of the conventional tools the quants used before the recession.

Let’s suppose that an insurance company faces the challenge to cover claims incurring in random time and size. The intensity of the process is $latex \lambda (t)=100t(1-t)$, $latex \lambda (0)=25$ and the process itself is (surprise!!!) Poisson. The claims are iid from a lognormal distribution with parameters $latex \mu=0.75t$ and $latex \sigma=1$, where $latex t$ is the intermediate time between occurence $latex i$ and $latex i-1$.

Some Mathematica code from my Graduate Simulation class is the following.

First we set up the intersity function and the generic vectors recoring Claims (Claimsv) and maximums (Maxv)

λ0 = 25; t0 = 1;
λ[t_] := 100 (t – t^2); Claimsv = {}; Maxv = {};

The main program is the following

Do[S = {}; t = 0; n = 0;
X = -Log[Random[]]/λ0; t = t + X;
While[t < t0, If[Random[] < λ[t]/λ0, n = n + 1; S = Append[S, t]];
X = -Log[Random[]]/λ0; t = t + X];
Claims = 0; S[[0]] = 0; Maxx = 0; Do[rr = Exp[
Random[NormalDistribution[0.75*(S[[i]] – S[[i - 1]]), 1]]];
If[rr > Maxx, Maxx = rr];
Claims = Claims + rr, {i, 1, n}] AppendTo[Claimsv, Claims];
AppendTo[Maxv, Maxx];, {10000}]

And plotting the histograms of the simulated distribution of the Claims and the maximum we nearly finished…

<< Histograms`

Histogram[Claims.v]
claimsv Histogram[Max.v]

maxv

to be continued…

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November 7, 2009

Confidence for the lognormal…

An intermediate exercise in Mathematical statistics is to construct a (100-$latex \alpha$)% confidence interval for the mean of a random variable distributed according to the log normal law ($latex X\tilde{\ }LN\left( \mu,\sigma _{{}}^{2}\right)$.)

The mean is

$latex E\left( X \right)=\exp \left( \mu +\frac{\sigma _{{}}^{2}}{2} \right)$.

Then

$latex \log \theta =\mu +\frac{\sigma _{{}}^{2}}{2}$.

The MVUE is

$latex T\left( Y \right)=\bar{Y}+\frac{s_{{}}^{2}}{2}$

, where $latex Y=\log X,{{s}^{2}}=\frac{1}{n-1}\sum{{{\left( Y-\bar{Y} \right)}^{2}}}$.

The variance of $latex T\left( Y \right)$ is

$latex Var\left\{ T\left( Y \right) \right\}=\frac{s_{{}}^{2}}{n}+\frac{s_{{}}^{4}}{2\left( n-1 \right)}$.

Then evidently

$latex \frac{\bar{Y}+\frac{s_{{}}^{2}}{2}-\log \theta }{\sqrt{\frac{s_{{}}^{2}}{n}+\frac{s_{{}}^{4}}{2\left( n-1 \right)}}}\xrightarrow{d}Z$

, where $latex Z\tilde{\ }N\left( 0,1 \right)$.

To sum thinks up the confidence interval for  mean of X fixing  $latex \alpha=.05$ is

$latex \left( \exp \left\{ \bar{Y}+\frac{s_{{}}^{2}}{2}-1.96\sqrt{\frac{s_{{}}^{2}}{n}+\frac{s_{{}}^{4}}{2\left( n-1 \right)}} \right\},\exp \left\{ \bar{Y}+\frac{s_{{}}^{2}}{2}+1.96\sqrt{\frac{s_{{}}^{2}}{n}+\frac{s_{{}}^{4}}{2\left( n-1 \right)}} \right\} \right)$.

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