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From LLN, we have $\text{plim} Y_n=\mu$. Now you have $W_n=Y^3_n$, what is the probability limit of $W_n$?

Please help me answer this and thank you!

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    $\begingroup$ Take a look at the continuous mapping theorem. $\endgroup$ Commented Feb 22, 2019 at 23:48
  • $\begingroup$ If you want to only find the probability limit, you can replace "plim" with "lim". Proving it involves probability theory. $\endgroup$
    – chan1142
    Commented Feb 23, 2019 at 8:35

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The comments have the answer. Let $\{X_n\}_{n=1}^{\infty}$ be a sequence of iid random variables with $E(X_1)=\mu$ and $Var(X_1)=\sigma^2$ with $\mu,\sigma^2<\infty$. Define $Y_n=\frac{1}{n}\sum_{i=1}^n X_i.$ By WLLN, $\text{plim}Y_n=\mu.$ Since $f(x)=x^3$ is a continuous function, by the continuous mapping theorem, $\text{plim}f(Y_n)=\text{plim}Y_n^3=f(\mu)=\mu^3.$

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