Thanks for the reply.
I am trying to valuate the following sum:
\sum_{i=0}^{n-1} C_i^{n-1} \; p^i (1-p)^{n-1-i} f(\frac{i}{n}) = c
where c is a real, positive constant, f is a decreasing function, and I am trying to evaluate p. I know that the end result when n → ∞ is p = f^{-1}...
Suppose that nx is binomially distributed: B((n-1)p, (n-1)p(1-p))
I wish to find the expected value of a function f(x), thus
\sum_{nx=0}^{n-1} B() f(x)
Assume that f() is non-linear, decreasing and continuous, f(x) = c is [0,1] to [0, ∞)
I want to show that the above sum converges to f(p)...
Thanks Stephen and Chiro, my question is not whether the example above satisfy WLLN, my question is about the Chebyshev's inequality. What can we conclude (if anything), if the RHS of the inequality is greater than 0 and smaller than ∞.
As for the example, I can prove that it does not satisfy...
I used tex for equations to make it look neater and easier to read (like how one would write in articles).
\sigma^2 is the variance of the sum of all \sigma_i divided by n (by independence, see below). So it is \text{Var} \left[ \frac{S_n}{n}\right] where S_n is \sum_{i=1}^{n} X_i...
Sorry about that, I keep omitting key information... you I am looking at sequences of independent random variables, so a simple example would be:
\Pr(X_i=±\sqrt{i})=0.5
Thanks for the latex help by the way.
Thank you Stephen, I am sorry that I didn't clarify. I meant for independent random variables, not iid. and by WLLN satisfied, I meant for a random variable, as n observations -> ∞, the sample mean -> μ
More formally, I can write:
Pr(|(sum(xi)/n - E(xi)|<epsilon)=1 for any epsilon>0 as n→∞
or...
Thank you very much Chiro, I have used that link but it doesn't really answer my question. Moreover, the WLLN in that link assumes i.i.d. where as I am only assuming independence. Thanks again
Sorry it's my first time posting so I am not sure if latex works here...
I have been trying to understand the proof of WLLN using Chebychev's inequalities, and here are my problems:
I know for the Strong Law of Large Numbers,
if n→∞, Ʃ(1 to n) σi ^2 / i^2 < ∞ => SLLN is satisfied...