Finance homework help - probability of share price

No. If the standard deviation were zero, and the return were therefore exactly 1.5% each month, the fund value would grow by 1.5% each month and therefore in a year it would grow by 19.56% (1.015^12 - 1).

I'm not really into all this standard deviation stuff, but I reckon if the SD is 1.2% it means that 50% (or some other proportion, depending on how the data are in fact distributed) of the time the return will be between 1.5%-1.2% and 1.5%+1.2%. So the worst case annual return for this middle range of returns should be 1.003^12-1 or 3.66% and the best case 1.027^12-1 or 37.67%. That kind of suggests that the annual SD would be half the difference, i.e. 17.005%, and the mean half way between the extremes, i.e. 20.665%.

This is probably all wrong, but I'm no statistician. Tim?

Well, £35 represents a growth from £30 of 16.67%. If we know that the expected annual mean is some 20.67%, and if we can assume the distribution to be symmetric, then we know the probability is 50% that the observed value will be greater than the mean. We want to know what the probability is of the value being between 16.67% and the mean, i.e. a 4%-wide band which therefore corresponds to 4/17 = 0.235 SDs. Your tables should be able to help you out there. The final probability required is then the sum of this and the aforementioned 50%.

Hello,

Many thanks for your reply and insight. I have the proper answers now and they make me think that the question itself was not as clear as it could have been.

For part a, the annualised yearly return was indeed 18% - the question was looking for the annualised *continuously compounded* rate which differs from the annual equivalent rate. An annual continuously compounded rate of 18% equates to an annually compounded rate of 19.56% as you stated, i.e. 1*e^0.18 = 1.1956. For the record, the annual continuously compounded standard deviation was indeed about 4.15%.

For part b, the probability that the share price is > £35 is apparently the equivalent of the probability of being greater than ln(35/30). So once you've found ln(35/30), you need to convert this number to a *standard* normal distribution probability, i.e. a z-score, and then look up the z-score in a standard normal distribution table or alternatively use Excel's NormSDist function.

To convert to a z-score, you subtract the annualised continuously compounded mean and then divide by the annualised continuously compounded standard deviation, so in this case as ln(35/30) = 0.1542, you convert this to a z-score as follows: (0.1542-0.18)/0.0415 and then apply NormSDist to this figure.

However, this isn't the end of the story! The figure obtained, 0.26 is actually the probability of being less than £35, so the actual ability we want is 1-0.26 = 0.74, i.e. 74%.

How c> >