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I have the following question:

Suppose that I have a sequence of random variables $X_n$ such that all moments exists and are finite. I have that $E[X_n]\to a$, where $a$ is a finite number and also $\text{Var}(X)\to 0$. We may also assume that a density with respect to the Lebesgue measure exists. Does this imply that $X_n\to_w \delta_a$? weakly?

Thanks a lot, any ideas on conditions are welcomed! This is true for the Gaussian distribution, but I wondered to what extend we can transfer it. Literature is also welcomed!

Davide Giraudo
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Liealgebrabach
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  • yes, suppose $Var(Y_n) \to 0$ and $E(Y_n) = 0$ then you have $E(|Y_n|^2) \to 0$ which means that $Y_n$ converges to the $0$ random variable in $L^2$, and this implies the convergence in distribution to the $0$ random variable (i.e. the pdf of $Y_n$ converges weakly to $\delta$). you can prove it by noticing that $P(|Y_n| > \epsilon) \to 0$ for every $\epsilon > 0$ because otherwise, $E(|Y_n|^2)$ would not tend $ \to 0$. see https://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_mean and https://en.wikipedia.org/wiki/Convergence_of_random_variables#Properties_4 – reuns Apr 10 '16 at 20:19
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    Notice that the reverse is not true. – leonbloy Jun 03 '16 at 13:12

1 Answers1

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Intuitively: the mean is almost constant, and the variance is smaller and smaller, hence we have are not too far away from the mean with a high probability.

More formally, we have $$\mathbb E\left[\left(X_n-a\right)^2\right]= \mathbb E\left[X_n^2\right]-2a\mathbb E\left[X_n\right]+a^2 =\operatorname{Var}\left(X_n\right)+\mathbb E\left[X_n\right]^2 +a^2-2a\mathbb E\left[X_n\right].$$ By assumption, $\operatorname{Var}\left(X_n\right)\to 0$ and $\mathbb E\left[X_n\right]^2 +a^2-2a\mathbb E\left[X_n\right]\to a^2+a^2-2a^2=0$, hence $X_n\to a$ in $\mathbb L^2$, hence in probability, hence in distribution.

Davide Giraudo
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