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Let $X = (X_i: i=1,\ldots,n)$ where $X_i \sim f$ i.i.d. such that $f$ is supported on $[0,\infty)$, $E(|X_i|)<\infty$ and $E(X_i)=\mu$. Consider $\tilde{X} = (X_i: X_i\leq a_n)$ for some deterministic sequence $a_n \to \infty$ and denote $m$ the number of elements in $\tilde{X}$.

By Law of large numbers we know that $\frac{1}{n}\sum_{i}^{n}X_i \to \mu$.

But, is it true that $\frac{1}{m}\sum_{i}^{m}\tilde{X}_i \to \mu$ ?

My idea is to try to use LLN again. Since $\tilde{X}_i \sim f_n$ with $f_n = \frac{f}{F(a_n)}$ and $F$ the c.d.f. of $X_i$, then $E(\tilde{X}_i)\leq E(X_i)<\infty$ and $E(\tilde{X}_i) = \frac{\int_0^{a_n}xf(x)dx}{F(a_n)}:= \mu_n$.

Then as $n\to\infty$ then $m\to\infty$ and $$ \left|\frac{1}{m}\sum_{i}^{m}\tilde{X}_i - \mu \right| \leq \left|\frac{1}{m}\sum_{i}^{m}\tilde{X}_i - \mu_n \right| + |\mu_n-\mu|$$ Therefore we conclude by using LLN on the first term and the second goes to 0 by definition and $a_n\to\infty$.

Is this correct? My main concern is that I'm trying to use LLN with a mean that depends on $n$. Is it valid?

Edit: Ok, this clearly seems to be wrong. What I should be able to do though is the following

$$ \left|\frac{1}{m}\sum_{i}^{m}\tilde{X}_i - \mu \right| \leq \left|\frac{1}{m}\sum_{i}^{m}\tilde{X}_i - \frac{1}{n}\sum_{i}^{n}X_i \right| + \left|\frac{1}{n}\sum_{i}^{n}X_i-\mu \right|$$ all I need to prove now is $$ \left|\frac{1}{m}\sum_{i}^{m}\tilde{X}_i - \frac{1}{n}\sum_{i}^{n}X_i \right| \to 0 $$ a.s. or in probability. How can I show that?

Gohan
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  • No, this argument is not valid. – Kavi Rama Murthy Nov 19 '19 at 23:42
  • is the statement still true though? if so how would you prove it? – Gohan Nov 19 '19 at 23:44
  • I think your $X$ and $\tilde X$ and your limits are too informally defined to create a useful proof. In particular, the sets are actually a function of $n$ but their notation does not reflect that. – orlp Nov 19 '19 at 23:47
  • Indeed, $\tilde{X}$ and $m$ depend on n, I just wanted to keep the notation simple. The idea is that for every $n$ I sample $n$ observations from $f$ and keep only the values that are less than $a_n$. Or equivalently, I sample from $f_n$, the truncated version of $f$. Does the average of these observations have the same limit as the non-truncated ones? – Gohan Nov 19 '19 at 23:57

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