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let us suppose we have following task : the husband is playing the internet game, in which he is winning or losing 1 dollar with equal probability, we should find a) mean value of this game and standard deviation

its seems that this game is a brownian motion, because at initial state husband has zero value

$W(0)=0$

and at probability of $p=0.5$ , he sometimes go up, sometimes go down, i found following link about brownian motion

brownian motion

but i could not find its connection to this problem, actualy mean value should be equal to $0$, because we have $x_1=1$ and $x_2=-1$, and expected value is equal to

$1*0.1+(-1)*0.5=0$

about standard deviation we would have

$1^2*0.5+(-1)^2*0.5-0^2=1$

from this we have $\sqrt{1} $

am i right?please help me , i am thinking about some distribution (binomial for instance) , we number of trials is equal to infinity, therefore

dato datuashvili
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    You might look at "random walks". – David K Oct 24 '17 at 18:14
  • mean is correct, what about standard deviation ? – dato datuashvili Oct 24 '17 at 18:20
  • i found https://web.stanford.edu/~dntse/classes/cs70_fall09/n15.pdf – dato datuashvili Oct 24 '17 at 18:23
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    Actually the variance comes out to $1$ (like you wrote before), not $2.$ The formula on the left of the $=$ is correct, it's just a matter of adding things up. (Notice that $0^2 = 0$; actually I don't see any reason to mention $0^2$ at all, since the terms $1^2 \times 1/2$ and $(-1)^2 \times 1/2$ cover all the possible outcomes of each game.) – David K Oct 24 '17 at 18:57
  • ouuu what a crazy mistake :D :D of course 0.5+0.5=1 dont pay attention – dato datuashvili Oct 24 '17 at 19:21
  • So I think you have the correct numbers now for the single game. If the husband keeps playing then the number of games grows without bound. If you can add to the question what it is you want to know about that series of games, maybe someone has an idea or a reference to suggest. – David K Oct 24 '17 at 20:13

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