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Say $X$ is a discrete random variable and its probability mass function $p_X(x)$ has finite number of mass points. I know that $\mathbb{E}(X^2)=0$ if and only if $\Pr(X=0)=1$.
However, I wonder if $\mathbb{E}[X^2]\downarrow0$, can I say that the probability mass function eventually has a probability mass point also converge to $0$? This seems intuitive to me but how can I prove it? Any suggestions? Thank you in advance for your time.

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The right question would presumably be on a sequence of random variables $\{X_n\}_{n=0}^\infty$ such that $E[X_n^2] \to 0$. This is equivalent to $E[(X_n - 0)^2] \to 0$, or that $X_n \to 0$ in the mean square sense. This implies convergence in probability to the constant zero, which implies convergence in distribution to the constant zero. This says that CDF's converge to a unit step function with a jump at $0$, which is precisely a point mass at zero.

(A slightly more elaborate related statement which you may want to show on your own is that if $X_n$ converges to $X$ in distribution and $X_n$ is integer valued, this is equivalent to $P(X_n = m) \to P(X=m)$ for every $n$; that is, you can state convergence in distribution in terms of PMF's rather than CDF's in this case. This is shown in Durrett's probability book. )

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  • $\begingroup$ Thanks batman, you are totally right! :) $\endgroup$ – aniki_aishiteru Sep 8 '14 at 10:19

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