Normal & log normal

With log-normal interest rate questions I'm confused about when do you/don't you have to
*Just times the mean & var you've been given by n and then stick them into the standardising formula
*log the mean & var you're given before putting them into the standardising formula
*stick the Mu value into eMu+0.5sigma^2 to calcuate a new Mean then put that as Mu into the standardising formula

or a combination of these

When I attempted Assignment Y1 Q1.12 I got it wrong and didn't quite understand the solution. So I've gone back to the course notes and seen the example on Pg 19. Yet the example seems to have done the same steps with the parameters as I did in the Assignment Y1 question.

So does anyone know a good website for explaining how to use the normal distribuion to calculate probabilities for a log-normal variable? I've looked on Wikipedia and it hasn't helped

Thanks

 

Hi Phil,

I'm guessing your talking about questions where your told that (1+i) has a lognormal distribution wth mean x and variance y.

My approach to these questions is to use the formuals on page 14 of the tables to find mu and (sigma)^2.

(Just a quick point to note. If your given that the mean is say 7% and variance is 3% you have to set E(1+i) = 1.07 to find mu and
var(1+i) = var(i) = .03)

then

(i+i) ~ logN(mu,(sigma)^2)

If interest rates are independent in each year then:

Sn ~ logN(n.mu,n.(sigma)^2)


If interest rates are not independent then I think its:

Sn ~ logN(n^2.mu, n^2.(sigma)^2)

Usually your then asked to find the probability that Sn or some initial investment will be greater than some final amount.
(eg: Probability that 1,000 invested at t=0 will be greater than 2,000 at t=5)


So you have found mu and (sigma)^2

and I'm assuming independence so

Sn ~ logN(n.mu,n.(sigma)^2)

(Sn also has a logNormal distribution with parameters mu = n.mu and
sigma^2 = n.(sigma^2)

Say your trying to find the probability that 1,000 will accumulate to greater that 2,000 after 5 years.

Then S5~ logN(5.mu,5.(sigma)^2)

you want to find

P(1,000S5 > 2,000)

(divide both sides by 1,000 and take the log of both sides)

P(ln(S5) > ln(2))

= 1 - P(log(S5) < ln(2))


Now you transform to the normal distribution using

x = (ln(2) - 5.mu) / Sqrt(5.(sigma)^2)

look up x in the tables on page 160 -161

then

P(ln(S5) > ln(2)) = 1 - Phi(x)


Hope this solves things for you!

 

Hi Joxer

Thanks for helping me

I think is the bit I'm confused about

Are you saying that if you're told (1+i) has a log-normal distribution with parameters mean = 0.075 and variance = 0.025^2, then from Pg 14

E(X) = exp(mu + 0.5(sigma)^2)
= exp(0.075 + 0.5(0.025^2)
= 1.078221

Var(X) = exp(2mu + sigma^2) (exp(sigma^2)-1)
= exp(0.15 + 0.025^2) (exp(0.025^2)-1)
= 0.000727

and then 1.078221, 0.000727 are the values you use in the standardising formula?


or do you mean that you create 2 equations
0.075 = exp(mu + 0.5(sigma)^2)
0.025^2 = exp(2mu + sigma^2) (exp(sigma^2)-1)

and solve simultaneously to find mu and sigma and they are what then go in the standardising formula?

 

Hi Phil,

You have to use simultaneous equations.

From page 14:

E(1+i) = exp[mu + (0.5)(sigma^2)]

Using the figures you gave

exp[mu + (0.5)(sigma^2)] = 1.075....................................................Eq 1

Rearrange equation 1 to get

ln(1.075) - (0.5)(sigma^2) = mu.......................................................Eq 2


Then from page 14:

Var(1 + i)= Var(i) = exp[2mu + (sigma^2)][exp(sigma^2) - 1]

Rearrange this to get:

Var(1 + i)= (exp[mu + (0.5)(sigma^2)])^2[exp(sigma^2) - 1].= .025^2...Eq 3

sub equation 1 into equation 3 to get:

Var(1 + i)= (1.075)^2[exp(sigma^2) - 1] = .025^2

exp(sigma^2)= 1 + (.025^2)/(1.075^2)

(sigma^2) = ln[1 + (.025^2)/(1.075^2)]

so (sigma^2) = .000540686

sub (sigma^2) back in to equation 2 to get:

mu = .072050318


This then gives

(1+i) ~ logN(0.072050318, 0.000540686)


Assuming i is independent in each year then

Sn ~ logN((n)x0.072050318, (n)x0.000540686)


So Sn also has a log normal distribution with parameters

mu =(n)x0.072050318 and (sigma^2) = (n)x0.000540686


Hope this makes things clearer!

Meant to say before, I had problems with chapter 15 too, couldn't find a good website but got my hands on the first 4 chapters of CT3 and that along with doing questions cleared up most of my problems.

 

Some clarification (hopefully)....

In questions we generally have that (1+i) has a lognormal distribution.

We write this (1+i) ~ logN(mu, sigma^2).

When we do this we call mu and sigma^2 the PARAMETERS of the lognormal distribution. mu is NOT the mean of the lognormal distribution (ie mu is not equal to the expected value of (1+i)) and sigma^2 is NOT the variance of the lognormal distribution (sigma^2 is not equal to var(1+i)).

There are two ways that we can be given the information to do our calculations and to specify the distribution fully.

1. We may be told the values of the parameters mu and sigma^2 explicitly in the question. Generally this means you will see "mu = XXXX and sigma^2= XXXX". Exactly as in the example on page 19 of Chapter 15 that Phil alludes to. If this happens, you can use these values directly, eg to standardise. (You may need to multiply by n if you're dealing wth an accumulation, as joxer states.)

2. We may be told information about the mean and variance of the interest rate distribution, ie we are given the mean and variance of i or (1+i). In this case we must set up the simultaneous equations as joxer has outlined (as we will know the values of E(1+i) and var(1+i)) and solve them for the parameters of the lognormal distribution mu and sigma^2. This is what happens in Y1 assignment Q1.12.

And they're the two ways I've seen the information presented.

Finally:

is true. But:

is not. If interest rates are not independent then it is:

Sn ~ logN(n.mu, n^2.sigma^2)

 

I'm not sure what to say. It's really just a statement of fact or you could treat it as a definition of a lognormal random variable.

If logX~N(mu,sigma^2), then this means that X~logN(mu,sigma^2).
Equally, if Y~logN(mu,sigma^2), then this means that logY~N(mu, sigma^2)

So, in essence, a lognormal RV is just a normal RV in disguise.

And note that log means natural log/ln/log base e, as it does everywhere in actuarial work.