What is average deviation from expectation?

  • Thread starter KFC
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  • #1
KFC
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Main Question or Discussion Point

In the book "The mathematic of Gambling", the author considers a fair coin with 50% getting head and 50% for tail. The expectation of such coin, of course, will be zero. Here is what I read from the text, it reads

Consider the fair game example mentioned earlier in the chapter (fair coin). In a series of any length, we have an expection of 0. In any such series it is possible to be ahead or behind. Your total profit or loss can be shown to have an average deviation from expectation of about [tex]\sqrt{N}[/tex].

Firstly, what is average deviation from expectation? How to get [tex]\sqrt{N}[/tex]? I am thinking the bionomial distribution with standard deviation [tex]\sqrt{N p (1-p)}[/tex] with p =0.5, it is about [tex]\sqrt{N}/2[/tex], is that what the author call average of deviation from expectation?
 

Answers and Replies

  • #2
uart
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Hi KFC, what they're referring to is that E(Y)=0 but E(Y^2)=N. Where Y is a sum of independant random variables each having mean=0 and varience=1.
 
  • #3
KFC
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Hi KFC, what they're referring to is that E(Y)=0 but E(Y^2)=N. Where Y is a sum of independant random variables each having mean=0 and varience=1.
Thanks uart. So why E(Y^2)=N? I know it said it is an approximation. Should the exact value be E(Y^2)=N/4 according to binomial distribution.
 
  • #4
uart
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For a sum of independent random variables the variance of the sum is equal to the sum of the variances. This true for all random variables no matter what the distribution, (binomial, normal, uniform etc), and its not an approximation. The only thing that is required is that they be independent.

For your binomial example the variance of each random variable is ( (1-0.5)^2 + (0-0.5)^2 )/2 = 0.25. So it follows that the variance of the sum of N such random variables is 0.25 N.

I assume that in the book you're reading they are normalizing the win/loss to one unit per game. That is, x=+1 for a win and x=-1 for a loss. So in this case the variance of X is ( (1-0)^2 + (-1-0)^2 )/2 = 1 and the variance of the sum of N such outcomes is equal to N.

BTW. Variance = E(X^2)
 
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  • #5
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"Just a little less than you expected."

I know it's not the math joke thread, but I had to do it.
 

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