Suppose you have $100,000 to invest in stocks. If you invest $1000 in any particular stock your profit will be $200, $100, $0, or -$100 with probabillity .25 each. There are 100 different stocks you can choose from, and they all behave independently of each other. Consider the two cases:
1) invest $100,000 in one stock
2) invest 1000 in each of the 100 stocks.
a)for case 1 find the probability that your profit will be $8000 or more.
b) do the same for case 2
Nov 7th 2009, 05:58 PM
You can brute-force the first part (obv the answer is .5), and the second part is probably looking for you to use the central limit theorem on the sum to get the answer. So, you'll calculate the expected profit, the variance, get a Z-Value for the required value of the sum to end up above 80,000(or equivalently the needed average to end up above 80), bust out the table, and find the probability of ending up above that.
Nov 9th 2009, 07:43 AM
I don't understand what you mean by brute force? how do you account for the fact that it is based on 1000$ investments? is it n=100 with each having the .25 probability? how do you do that with the part a - do you think of all the ways it could sum to at least 8000?
Thank you for helping!!!
Nov 9th 2009, 08:18 AM
If you invest 100,000 in one stock, you are investing 1,000 in it 100 times. This is the same stock; it will only have one return. If that stock returns $100 per $1000 or $200 per $1000, your total profit will be $10,000 and $20,000 respectively. Both of those exceede 8,000, while the other two outcomes don't. That is, I think, what they intended for you to do, since they appear to be wanting to impress on you the principal of diversification.