Lognormally distributed

#1
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

TS Contributor
#2
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-normal_distribution#Mean_and_standard_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.
 
#3
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-normal_distribution#Mean_and_standard_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

TS Contributor
#4
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.