Lognormally distributed

**Sigrino** · 01-17-2011 04:31 PM

Help please I am stuck.

The rate of return on Your portfolio, R, has a mean value of 1% and a standard deviation of 5%. Suppose that (1 + R) is lognormally distributed.

1) Calculate at what level of rate of return y, the probability that R is less than or equal to y is equal to 10%.

2) Calculate the probability that R is greater than 10%.

P.S. I undarstand that the shape of a lognormall distribution is skeewed to the left. But do not understand how to solve this. Is there a trick to simplify this to a normal distribution? If so please help and explain it to me. Thank you.

**BGM** · 01-17-2011 11:43 PM

Several things you may need to know:

1. If $1 + R \sim \mathrm{log}\mathcal{N}(\mu, \sigma^2)$ ,
then $\ln(1 + R) \sim \mathcal{N}(\mu, \sigma^2)$

http://en.wikipedia.org/wiki/Log-nor...dard_deviation

2. If $1 + R \sim \mathrm{log}\mathcal{N}(\mu, \sigma^2)$ ,
then $E[1 + R] = \exp\left\{\mu + \frac {\sigma^2} {2}\right\}, Var[1 + R] = (\exp\{\sigma^2\} - 1)\exp\{2\mu + \sigma^2\}$

3. Since you know the mean and variance of the log normal distribution,
you may solve the parameters $\mu, \sigma^2$ and thus
you can convert any problem about the probability of a lognormal random
variable into a problem about the normal random variable.

**Sigrino** · 01-20-2011 07:58 AM

Originally Posted by BGM

Several things you may need to know:

1. If $1 + R \sim \mathrm{log}\mathcal{N}(\mu, \sigma^2)$ ,
then $\ln(1 + R) \sim \mathcal{N}(\mu, \sigma^2)$

http://en.wikipedia.org/wiki/Log-nor...dard_deviation

2. If $1 + R \sim \mathrm{log}\mathcal{N}(\mu, \sigma^2)$ ,
then $E[1 + R] = \exp\left\{\mu + \frac {\sigma^2} {2}\right\}, Var[1 + R] = (\exp\{\sigma^2\} - 1)\exp\{2\mu + \sigma^2\}$

3. Since you know the mean and variance of the log normal distribution,
you may solve the parameters $\mu, \sigma^2$ and thus
you can convert any problem about the probability of a lognormal random
variable into a problem about the normal random variable.

I solved the equestion for E[1+R] and Var[1+R], by inputing mu=0.01 and sigma=0.05.

So I got
E[1+R]=1.012578452
Var[1+R]=0.002560086

But I don't understand how to go from here. I am asked to find (R<y)=10%

But I do not have E[R] and Var[R] values, and I am not sure how I can convert

E[1+R] in to E[R], and Var[1+R] into Var[R].

Maybe I do not full understand somthing, if u see where I went wrong in my thought process, please point it out. Thank you for the help. This question is bugging me very much.

**BGM** · 01-21-2011 02:46 PM

The parameters I typed $\mu, \sigma^2$ are not the mean
and variance of $R$ .
They are the mean and variance of $\ln(1 + R)$

So the first thing you should know

$E[1 + R] = 1 + E[R] = 1 + 1\%, Var[1 + R] = Var[R] = (5\%)^2$

Next, solve for the $\mu, \sigma^2$ which is a two equations
with two unknowns. It is easy. You can either look for the solution at the
wikipedia, or consider $E[1 + R]^2$

After obtaining the parameters, note that $\ln$ function is a
strictly increasing function, and the inequality will preserve. So the required
probability $\Pr\{R < y\} = 10\%$ and $\Pr\{R > 10\%\}$ is easy once you have access to the normal CDF.

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