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lord12
Topic Author
Posts: 33
Joined: February 1st, 2007, 12:42 am

### probability of stock going up or down

r = 0.1 u = 1.2 d=.9. John thinks the stock will go up and makes a bet with Jerry. If S1>S0, John will $100 from Jerry. If S0<S1, John will give$100 to Jerry. Who has the advantage?

quantmeh
Posts: 5974
Joined: April 6th, 2007, 1:39 pm

### probability of stock going up or down

looks like it's John.risk-neutral p is (1.1-0.9)/(1.2-0.9) = 2/3, so his agreement with Jerry is worth {100(2/3)-100(1/3)}/1.1=100/3.3 ~ 30, while he entered into it paying nothing.

Zedr0n
Posts: 244
Joined: April 6th, 2007, 5:07 am

### probability of stock going up or down

Hm... What are the u and d here? If I assume that it's log-normal drift and volatility thenThe bet is equivalent to going long one digitall call option and going short one digital put optionSo the arbitrage free price under log-normal(Black-Scholes) model isIn our case 0.9*0.9/2 - 0.1 > 0 => N(...) > 0.5 => 1 - 2N(...) < 0 => price is less than zero, so John should be paid for entering such a bet by Jerry, while Jerry doesn't so Jerry has the advantage

Haron
Posts: 4
Joined: February 22nd, 2007, 4:36 pm

QuoteOriginally posted by: lord12... If S1>S0, John will $100 from Jerry. If S0<S1... - 8))))Noone has advantage: John will trade his 100 for Jerry's 100 because S1>S0 and S0<S1 will happen at the same time letiand Posts: 11 Joined: April 3rd, 2007, 1:58 am ### probability of stock going up or down QuoteOriginally posted by: lord12r = 0.1 u = 1.2 d=.9. John thinks the stock will go up and makes a bet with Jerry. If S1>S0, John will$100 from Jerry. If S0<S1, John will give \$100 to Jerry. Who has the advantage?An idea is the following: five cases for each guy. I think theycan hedge out the risk which depends on the situation of option at that time. The clue is :1) They can not hedge out risk if only hold/sell the stock...In the side of Jerry:1. Jerry can buy X stock(Let us suppose the price is 1) at time 0 , if stock is up, he will get X*1.2-100 ,if stock is down, he will get X*0.9+100. So, he need X*1.2-100>(1+r)*x=1.1X and X*0.9+100>1.1Xto hedge out any risk, 0.1X>100 and 100>0.2X hold at the same time. it is impossible .In the side of John, he need X*1.2+100>1.1X and X*0.9-100>1.1X So X>-1000 and -100>0.2X (X<-500). We get-500<X<-1000! it is impossible if John buy stock. 2) But it seems promising if they can use call/put option to hedge out the risk. Eg: John can sell the call option at time 0 and the strike is K. the call option price per stock is then (1.2-K)*2/3.Suppose he sell X call option to his client, the total money at time 1 is if stock is up, he will get (1+r)*X*(1.2-K)*2/3-(1.2-K)*X+100. So if 100>(1.2-K)*0.27X, he will win 100- (1.2-K)*0.27X bucks.Notice we need X<100/((1.2-K)*0.27) If stock is down, he will get (1+r)*X*(1.2-K)*2/3+(K-0.9)*X-100=> 0.73*(1.2-K)X+KX-0.9X-100,we need 0.23K*X-0.02X-100>0 to get profit, so he needs X>100/(0.23K-0.02) .In this way, we know if 100/((1.2-K)*0.27)>100/(0.23K-0.02)=> 0.23K-0.02>0.32-0.27K=>0.5K>0.34 => K>0.68=> Always hold if we know u=1.2 d=0.9.In this example, if John is a banker, he can bet with Jerry as he can hedge out therisk completely! Any idea?
Last edited by letiand on April 10th, 2007, 10:00 pm, edited 1 time in total.

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