September 2012 Q7 part i)

Can anyone explain what the examiners' report means by "standard interpolation"?
I would have approached this question by calculating two values for the call option price, first using sigma=0.2, and then increasing or decreasing appropriately, and then interpolating. But mark allocation for this seems to be very stingy - only 2 marks! Implying that I should be able to do all this in around 3.6 minutes!!!
In previous papers there was usually around 5 marks for this type of question.
Is there a quicker way of doing this? Any ideas?
Thanks!

 

not to mention that the answer is apparently a ridiculous 436%! could there have been a mistake in the question?

 

i dont think there's a quicker way...but you can guess that the volatility is quite high by the ratio So/Co (the call option is out-of-the-money, yet its price is very high. so its time value is high. it must be due to the high volatility).

if you spend too much time interpolating for the variance, its probably better for you to avoid questions like this in the exam.

i do agree that the marking for this question is a bit mean for the time one has to spend on it.

 

Dazed and confused,

Your method is correct, as is your starting point. I always suggest starting with 20%, unless there's an explicit suggestion otherwise in the question.

Bluetails' comment about the call price being so (relatively) high is an indication that sigma is likely to be high here. For an out-of-money call to have a price of $0.86 when the share price is only $1.20 is very unusual. However, I don't imagine that anyone sitting the exam thought it would turn out to be over 400%!!

I suspect that there was a typo in the question and that the call price should actually have been 0.86 cents, as this corresponds to a volatility of 16%.

Regardless, I agree that 2 marks = 3.6 minutes doesn't seem very long to do the interpolation.

 

Thanks Graham,

It's comforting that you agree.

The examiners comment "there was evidence of candidates spending a significant amount of time in part (i)" seems to suggest that you would be expected to do the interpolation at lighting speed. I guess if d1 and d2 and sigma itself turned out to be nice numbers, then maybe you could do it in a couple of minutes. But with the the correct answer actually being well over 100% (which would usually set my alarm bells ringing that I've probably done something wrong), I think this was very mean.

Let's hope there's no typos tomorrow!