# Stock forward price argument

Hi I am strangling to understand where is the mistake with the following strategy. Can anyone help me with the following argument?

Assuming a stock price follows geometric Brownian motion then the expected value, under the physical measure, is $S\exp(μt)$. If the forward price is $S\exp(rt)$ then if I always long the forward I will be profitable in the long run ( i understand there is a risk envolve and that this difference it can be explain by the market price of risk. I also understand that if the price of the forward is $S\exp(μt)$ there will be arbitrage by borrowing money buying the stock and selling it at the forward price.) But still under those assumptions if the price of forward is $S\exp(rt)$ by entering long I will be profitable in the long run.

Can anyone point out where my mistake is?

• Do you want to know what is wrong with your idea, or if there is something wrong with your idea given the constraints you have imposed upon the problem. Jul 12 '18 at 16:38
• You need to clarify your question, what are you trying to achieve precisely? Anyway, to help clarify things for you, the price of the forward is given by a no arbitrage argument. Buying the stock and holding it until $t$ while selling the forward is not an arbitrage, you have a non-zero probability to lose money. Jul 13 '18 at 11:35