Let $X_n$ a random variable of same law

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If the variance of the random variable \(X_n\) approaches zero as \(n\) increases, it implies that the random variables are converging in probability to a constant \(C\). Consequently, the expected value \(E(X_n)\) will converge to \(C\), and the expected value of the square \(E(X_n^2)\) will converge to \(C^2\). A suggested approach to demonstrate this involves redefining the variable as \(z = x - E(x)\) and proving that if \(E(z^2) = 0\), then \(z\) must have a discrete probability density with \(P(z=0) = 1\). This indicates that the random variables are effectively clustering around the constant \(C\). The discussion emphasizes the relationship between variance, expected values, and convergence in probability.
Feynman
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:confused: Hello

I have a simple question :
Let $X_n$ a random variable of same law
If $V(X_n)\longrightarrow 0$ when $n\longrightarrow +\infty$
How schow that : $E(X_n)\longrightarrow C$ and $E(X_{n}^{2}\longrightarrow C^2$ and C is a constant?
Thanks
 
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Hello

<br /> <br /> I have a simple question :<br /> Let $X_n$ a random variable of same law<br /> If $V(X_n)\longrightarrow 0$ when $n\longrightarrow +\infty$<br /> How schow that : $E(X_n)\longrightarrow C$ and $E(X_{n}^{2}\longrightarrow C^2$ and C is a constant?<br /> Thanks<br />
 
Feynman, LaTex is down at the moment.
 
How far have you got with this one Feynman?

I suggest you let z = x - E(x) and then prove that E( z^2 ) = 0 implies that Z has a discrete probability denstiy with P(z=0) = 1.
 
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