Dominated Convergence and Truncation Arguments for Discrete Variables
Questions that involve truncating a discrete random variable (e.g., X·1_{|X|≤C}), showing convergence of truncated expectations, or bounding approximation errors from truncation.
Let $X$ be a discrete random variable with finite expectation. Show that $$\mathbb{E}\left(X \mathbb{1}_{|X| \leqslant C}\right) \xrightarrow{C \rightarrow +\infty} \mathbb{E}(X).$$
For every $(i,j) \in (\mathbb{N}^{\star})^{2}$ and for every $C > 0$, we set $\sigma_{ij}(C) = \sqrt{\mathbb{V}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)}$. Deduce that $$\lim_{C \rightarrow +\infty} \sigma_{ij}(C) = 1$$
For every $(i,j) \in (\mathbb{N}^{\star})^{2}$ and for every $C > 0$, we set $\sigma_{ij}(C) = \sqrt{\mathbb{V}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)}$. If $\sigma_{ij}(C) \neq 0$, we set $$\widehat{X}_{ij}(C) = \frac{1}{\sigma_{ij}(C)}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C} - \mathbb{E}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)\right).$$ Justify that, for $C$ large enough, the variables $\widehat{X}_{ij}(C)$ are well defined and that they are then bounded, centered, of variance 1 and that they are mutually independent for $1 \leqslant i \leqslant j$.
For every $(i,j) \in (\mathbb{N}^{\star})^{2}$ and for every $C > 0$, we set $\sigma_{ij}(C) = \sqrt{\mathbb{V}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)}$. If $\sigma_{ij}(C) \neq 0$, we set $$\widehat{X}_{ij}(C) = \frac{1}{\sigma_{ij}(C)}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C} - \mathbb{E}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)\right).$$ Show that $$X_{ij} - \widehat{X}_{ij}(C) = \left(1 - \frac{1}{\sigma_{ij}(C)}\right) X_{ij} + \frac{1}{\sigma_{ij}(C)}\left(X_{ij} \mathbb{1}_{|X_{ij}| > C} - \mathbb{E}\left(X_{ij} \mathbb{1}_{|X_{ij}| > C}\right)\right).$$
For every $(i,j) \in (\mathbb{N}^{\star})^{2}$ and for every $C > 0$, we set $\sigma_{ij}(C) = \sqrt{\mathbb{V}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)}$. If $\sigma_{ij}(C) \neq 0$, we set $$\widehat{X}_{ij}(C) = \frac{1}{\sigma_{ij}(C)}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C} - \mathbb{E}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)\right).$$ Show that $$\lim_{C \rightarrow +\infty} \mathbb{E}\left(\left(X_{ij} - \widehat{X}_{ij}(C)\right)^{2}\right) = 0.$$