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I am self-studying and I am working on the following problem:

enter image description here

My solution is different and I'm arriving at a different answer:

The parameters of the lognormal random variable $S_t/S_0$ are: $$m = \mu t = (\alpha - \delta - 0.5\sigma^2)t$$ and $$v = \sigma\sqrt{t},$$ where $\alpha$ is the stock's continuously compounded rate of return, $\sigma$ is the stock's annual volatility, and $t = \frac{1}{250}$.

Then we have $m = (\alpha - 0.5\sigma^2)/250$ and $v = \sigma\sqrt{1/250}$.

Therefore:

$40e^{m + z_1v} = 40e^{(\alpha - 0.5\sigma^2)/250 + 1.18\sigma\sqrt{1/250}} = 40.866$

and

$40.866e^{m + z_2v} = 40e^{(\alpha - 0.5\sigma^2)/250 - 0.53\sigma\sqrt{1/250}} = 40.519.$

This gives us the system of equations:

$(\alpha - 0.5\sigma^2)/250 + 1.18\sigma\sqrt{1/250} = \ln(40.886/40)$

and

$(\alpha - 0.5\sigma^2)/250 - 0.53\sigma\sqrt{1/250} = \ln(40.519/40.866).$

Solving yields $\alpha = 0.2666063$ and $\sigma = 0.281421201$.

It appears to me that the author is starting with the lognormal parameters already in terms of 1 day, and then converting to the annual return/volatility at the end. I am starting with the annual lognormal parameters, converting to daily, and then solving for the annual return/volatility.

If my reasoning is correct, I don't see why my solution wouldn't match the one using the author's method.

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you got a typo. It should be 40.886 in your last equation. Then $\sigma$ should match.

Also, If $\alpha$ means annualized log return, it should be

$\mu\,t = \alpha - \frac 1 2\sigma^2\,t$

So in your last two equations, the first term should be

$\frac \alpha {250} - \frac 1 2 \sigma^2$

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  • $\begingroup$ Just by correcting 40.866 to 40.886, I arrive at the same answer for $\alpha$ and $\sigma$ as the author, without having to adjust the last two equations. Would annualized return not be $\alpha = \mu t = (\alpha - 0.5\sigma^2)*t = (\alpha - 0.5\sigma^2)\cdot\frac{1}{250}$? I would think that both $\alpha$ and $0.5\sigma^2$ need a factor of $\frac{1}{250}$ in my last two equations. $\endgroup$ – user2521987 Dec 7 '16 at 15:28

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