The Reading 36 CFAI text states that “It can be shown that for a lognormal distribution the standard deviation of the one-year rate is equal to i0σ.” That is, standard deviation of the 1-year rate = 1-year spot rate x sigma.

The text states that sigma = “assumed volatility of the one-year rate”. Therefore, doesn’t that mean that sigma = the standard deviation of the 1-year rate?

I can’t understand how the standard deviation is derived from itself. How can something be a function of itself? I am at a loss.

a variable is lognormal distributed if log(variable) is normally distributed.

So log(X) = N(Mu, sigma).

so you have sigma available to you already. What is changing is the time period – so 1 year spot rate * sigma of the basic (distribution) is the standard deviation of the 1 year lognormal variable.

So, when they say that adjacent possible outcomes in the binomial tree are two standard deviations apart, are they referring to the standard deviation of the normally distributed variable or the standard deviation of the lognormally distributed variable?

Thanks! From what you both have said and my own google searching, I gather that the interest rates are assumed to be lognormally distributed and that interest rate changes are normally distributed (due to the up and down moves being computed using logs). Is that correct?

Also, I’m still trying to understand why the standard deviation of the lognormal variable is 1-yr interest rate * sigma. CFAI text gives these workings:

e^2sigma = approx. 1 + 2sigma

therefore standard deviation = (r * e^2sigma - r)/2, which is approximately equal to:

(r + (r * 2sigma) - r)/2

= r * sigma

Thanks cpk123 for explaining that this is because what’s changing is the time period. I still don’t really understand why it is derived like this, though. I suppose these workings do make sense, if I accept that the difference between the two adjacent paths is two standard deviations, but I would like to understand the workings without it being dependent on this fact (as in, I would like to be able to understand the workings as its own self-contained explanation), if you know what I mean.

Thanks so much for being patient and explaining all this to me, I really do appreciate it! I’m the kind of person who likes to understand why and what they’re doing.