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Let $(X_{jn})_{1\leq j \leq n}$, $n\in \mathbb N$, be a triangular array of random vectors in $\mathbb R^d$ (the $X_{jn}$ are understood to be independent in $j$ for fixed $n$.). We say that the triangular array is null if $X_{jn} \overset{p}{\to} 0$, as $n \to \infty$, uniformly in $j$, i.e.,: \begin{equation} \lim_{n \to \infty} \max_{1\leq j\leq n} P( \, |X_{jn}|> \epsilon \, )=0, \quad (\forall \, \epsilon>0). \end{equation} For any random vector $X$ with distribution $\mu$, we introduce an associated compound Poisson random vector $X^{\tilde{}}$ (Note the tilde, please) with characteristic measure $\mu$, i.e.: \begin{equation} \log \varphi_{X^{\tilde{}}}(u) = \int_{\mathbb R^d} (e^{iux}-1)\mu (dx), \quad u \in \mathbb R^d. \end{equation} For a triangular array $(X_{jn})_{1\leq j \leq n}$, the corresponding compound Poisson vectors $(X_{jn}^{\tilde{}})_{1\leq j \leq n}$ are again assumed to have row-wise independent entries.

By $X_n \overset{d}{\sim} Y_n$ we mean that, if either side converges in distribution along a sub-sequence, then so does the other along the same sequence.

We can show the proposition about compound Poisson approximation: Let $(X_{jn})_{1\leq j \leq n}$, $n\in \mathbb N$, be a null array of random vectors in $\mathbb R^d$. Fix $h > 0$, and define: $$b_{jn}= E(X_{jn}\,;\,|X_{jn}|\leq h).$$ Then $$\sum_j X_{jn} \overset{d}{\sim} \sum_j \left\{ (X_{jn} - b_{jn})^{\tilde{}} +b_{jn} \right\}$$

(This the Proposition 7.11 from Foundations of Modern Probability, Third edition)

Question:

Suppose $E[X_{jn}]=0$ for all $j$ and $n$. How to show this?: $$\sum_j X_{jn} \overset{d}{\sim} \sum_j X_{jn}^{\tilde{}} $$ Intuitively, when $h \to \infty$, we have $b_{jn}\to 0$.

I do not know if $E[X_{jn}]=0$ is sufficient. If this condition is not sufficient, I also have that: \begin{equation}\label{I}\tag{I} \sum_{j=1}^n \mathbf{v}(X_{jn})\leq C< \infty,\quad \forall \, n \end{equation} where $X_{jn}=(X_{jn_{1}},..., X_{jn_{d}})$ and: $$\mathbf{v}(X_{jn}):=\hbox{trace} \left(E\left[X_{jn}X_{jn}^{\,\,'}\right]\right)= \sum_{k=1}^d \hbox{var}(X_{jn_{k}})$$ Note that $X_{jn_{k}}$ is unidemsional, $E[X_{jn_{k}}]=0$ and $\hbox{var}(X_{jn_{k}})= E[(X_{jn_{k}})^2]$.

Let $(X_{jn})_{1\leq j \leq n}$, $n\in \mathbb N$, be a triangular array of random vectors in $\mathbb R^d$ (the $X_{jn}$ are understood to be independent in $j$ for fixed $n$.). We say that the triangular array is null if $X_{jn} \overset{p}{\to} 0$, as $n \to \infty$, uniformly in $j$, i.e.,: \begin{equation} \lim_{n \to \infty} \max_{1\leq j\leq n} P( \, |X_{jn}|> \epsilon \, )=0, \quad (\forall \, \epsilon>0). \end{equation} For any random vector $X$ with distribution $\mu$, we introduce an associated compound Poisson random vector $X^{\tilde{}}$ (Note the tilde, please) with characteristic measure $\mu$, i.e.: \begin{equation} \log \varphi_{X^{\tilde{}}}(u) = \int_{\mathbb R^d} (e^{iux}-1)\mu (dx), \quad u \in \mathbb R^d. \end{equation} For a triangular array $(X_{jn})_{1\leq j \leq n}$, the corresponding compound Poisson vectors $(X_{jn}^{\tilde{}})_{1\leq j \leq n}$ are again assumed to have row-wise independent entries.

By $X_n \overset{d}{\sim} Y_n$ we mean that, if either side converges in distribution along a sub-sequence, then so does the other along the same sequence.

We can show the proposition about compound Poisson approximation: Let $(X_{jn})_{1\leq j \leq n}$, $n\in \mathbb N$, be a null array of random vectors in $\mathbb R^d$. Fix $h > 0$, and define: \begin{equation}\label{k1}\tag{K1} b_{jn}= E(X_{jn}\,;\,|X_{jn}|\leq h). \end{equation} Then \begin{equation}\label{k2}\tag{K2} \sum_j X_{jn} \overset{d}{\sim} \sum_j \left\{ (X_{jn} - b_{jn})^{\tilde{}} +b_{jn} \right\} \end{equation} (This the Proposition 7.11 from Foundations of Modern Probability, Third edition)

Update

Allow me to rephrase my question. We say that a Infinite Divisible r. vector has the levy kintchine representations if its characteristic function is given by: $$\varphi_X(u)= \exp\left\{i u'b + \frac{1}{2}u'au + \int_{\mathbb R^d} \left[e^{iu'x}-1 - i u'x c(x)\right] d\nu(x) \right\} $$ where $c(x)$ is a integrable function. We denote this as $X \sim (b_c, a, \nu)_c$. In the case above, we have that $X^{\tilde{}} \sim (0,0,\mu)_0$.

Now, we can change the truncation function $c(x)$ by other $h(x)$. If $X \sim (b_c, a, \nu)_c$, we have that $$X \sim (b_h, a, \nu)_h, \quad b_h = b_c + \int_{\mathbb R^d}x [ h(x)- c(x)]d\nu(x)$$

Question:

Given a Null triangular array $(X_{jn})_{1\leq j \leq n}$ with $X_{jn} \sim \mu_{jn}$. Suppose $E[X_{jn}]=0$ and we also have that: \begin{equation}\label{I}\tag{I} \sum_{j=1}^n \mathbf{v}(X_{jn})\leq C< \infty,\quad \forall \, n \end{equation} where $X_{jn}=(X_{jn_{1}},..., X_{jn_{d}})$ and: $$\mathbf{v}(X_{jn}):=\hbox{trace} \left(E\left[X_{jn}X_{jn}^{\,\,'}\right]\right)= \sum_{k=1}^d \hbox{var}(X_{jn_{k}})$$ Note that $X_{jn_{k}}$ is unidemsional, $E[X_{jn_{k}}]=0$ and $\hbox{var}(X_{jn_{k}})= E[(X_{jn_{k}})^2]$.

Define $S_n' \sim (0,0,\mu_n)_h$ with $h(x)\equiv 1$ and $$\mu_n := \sum_{j=1}^n \mu_{jn}$$ So, how to show, using (\ref{k1}) and (\ref{k2}), that: $$S_n := \sum_j X_{jn} \overset{d}{\sim} S_n' $$

Let $(X_{jn})_{1\leq j \leq n}$, $n\in \mathbb N$, be a triangular array of random vectors in $\mathbb R^d$ (the $X_{jn}$ are understood to be independent in $j$ for fixed $n$.). We say that the triangular array is null if $X_{jn} \overset{p}{\to} 0$, as $n \to \infty$, uniformly in $j$, i.e.,: \begin{equation} \lim_{n \to \infty} \max_{1\leq j\leq n} P( \, |X_{jn}|> \epsilon \, )=0, \quad (\forall \, \epsilon>0). \end{equation} For any random vector $X$ with distribution $\mu$, we introduce an associated compound Poisson random vector $X^{\tilde{}}$ (Note the tilde, please) with characteristic measure $\mu$, i.e.: \begin{equation} \log \varphi_{X^{\tilde{}}}(u) = \int_{\mathbb R^d} (e^{iux}-1)\mu (dx), \quad u \in \mathbb R^d. \end{equation} For a triangular array $(X_{jn})_{1\leq j \leq n}$, the corresponding compound Poisson vectors $(X_{jn}^{\tilde{}})_{1\leq j \leq n}$ are again assumed to have row-wise independent entries.

By $X_n \overset{d}{\sim} Y_n$ we mean that, if either side converges in distribution along a sub-sequence, then so does the other along the same sequence.

We can show the proposition about compound Poisson approximation: Let $(X_{jn})_{1\leq j \leq n}$, $n\in \mathbb N$, be a null array of random vectors in $\mathbb R^d$. Fix $h > 0$, and define: $$b_{jn}= E(X_{jn}\,;\,|X_{jn}|\leq h).$$ Then $$\sum_j X_{jn} \overset{d}{\sim} \sum_j \left\{ (X_{jn} - b_{jn})^{\tilde{}} +b_{jn} \right\}$$

(This the Proposition 7.11 from Foundations of Modern Probability, Third edition)

Question:

Suppose $E[X_{jn}]=0$ for all $j$ and $n$. How to show this?: $$\sum_j X_{jn} \overset{d}{\sim} \sum_j X_{jn}^{\tilde{}} $$ Intuitively, when $h \to \infty$, we have $b_{jn}\to 0$.

I do not know if $E[X_{jn}]=0$ is sufficient. If this condition is not sufficient, I also have that: \begin{equation}\label{I}\tag{I} \sum_{j=1}^n v(X_{jn})\leq C< \infty,\quad \forall \, n \end{equation}

where $v(X_{jn}):=\hbox{trace} \left(E(X_{jn}X_{jn}^{\,\,'})\right)= \sum_{k=1}^d var(X_{jn_{k}})$ and $X_{jn}=(X_{jn_{1}},..., X_{jn_{d}})$. Note that $X_{jn_{k}}$ is unidemsional, $E[X_{jn_{k}}]=0$ and $var(X_{jn_{k}})= E[(X_{jn_{k}})^2]$.

My attempt

Note that the compound Poisson random vector $X_{jn}$ is such that $X_{jn}^{\tilde{}}= \sum_{\kappa=1}^N X_{\kappa;jn}$, where $(X_{\kappa;jn})_{\kappa=1}^\infty$ are iid copies of $X_{jn}$ and $N \sim \hbox{Poisson}(1)$. So $$(X_{jn} - b_{jn})^{\tilde{}}= X_{jn}^{\tilde{}} - N b_{jn}$$ Now, agregating we have: $$ \begin{aligned} \sum_j \left\{ (X_{jn} - b_{jn})^{\tilde{}} +b_{jn} \right\}&=\sum_j (X_{jn} - b_{jn})^{\tilde{}} +\sum_j b_{jn}\\ &=\sum_j X_{jn}^{\tilde{}} - (N-1)\sum_j b_{jn} \end{aligned} $$ Thus, it remains to show that $\sum_j b_{jn}=0$ as $h \to \infty$.

For this, I will do $h \in \mathbb N$. Since \begin{equation}\label{II}\tag{II} X^h_{jn}:= X_{jn} \,\mathbf{1}_{[|X|_{jn}< h]} \overset{a.s}{\longrightarrow} X_{jn}, \quad(h \to \infty) \end{equation} my first attempt was to use the Dominated Convergence Theorem for the random vectors for $X^h_{jn}$. But I can't find $Y$ with $E[Y]< \infty$ such that $|X^h_{jn}|\leq |Y|$, for all $h$. So I think that (\ref{I}) is necessary. Using this condition I will try to show uniform integrability and conclude.

First, suppose that $d=1$. Defining $S_h := \sum_{j=1}^n X^h_{jn}$, $h \in \mathbb N$. Remember that $X_{in}$ and $X_{jn}$ are independent. So, using this result, we have that (i) $X_{in}^h$ and $X_{jn}^h$ are independent. Moreover, note that (ii) $|X^h_{jn}|\leq |X_{jn}|$. Thus, we have:
$$E[S_h^2]\overset{(i)}{=} \sum_{j} E[(X^h_{jn})^2] \overset{(ii)}{\leq} \sum_{j} E[X_{jn}^2]\overset{(\ref{I})}{\leq}C$$ This implies that: \begin{equation}\label{III}\tag{III} \sup_{h }E [S_h^2]< \infty \end{equation} By (\ref{II}), we have that $S_h = \sum_{j=1}^n X^h_{jn}\overset{d}{\longrightarrow} \sum_{j=1}^n X_{jn}$, as $h \to \infty$. This together with (\ref{III}), we conclude that $$\sum_{j=1}^n b_{jn}= E \left[S_h \right] \longrightarrow E \left[\sum_{j=1}^n X_{jn}\right]=0, \quad (h \to \infty).$$

Well, I would like your help to know if:

a) For the case $d=1$ I am not making any big mistakes. Is my strategy right?

b) If the case $d=1$ is correct, how can we generalize to higher dimensions? Honestly, I don't know if there is any uniform integrability result using the variance for vectors that I defined above.

Let $(X_{jn})_{1\leq j \leq n}$, $n\in \mathbb N$, be a triangular array of random vectors in $\mathbb R^d$ (the $X_{jn}$ are understood to be independent in $j$ for fixed $n$.). We say that the triangular array is null if $X_{jn} \overset{p}{\to} 0$, as $n \to \infty$, uniformly in $j$, i.e.,: \begin{equation} \lim_{n \to \infty} \max_{1\leq j\leq n} P( \, |X_{jn}|> \epsilon \, )=0, \quad (\forall \, \epsilon>0). \end{equation} For any random vector $X$ with distribution $\mu$, we introduce an associated compound Poisson random vector $X^{\tilde{}}$ (Note the tilde, please) with characteristic measure $\mu$, i.e.: \begin{equation} \log \varphi_{X^{\tilde{}}}(u) = \int_{\mathbb R^d} (e^{iux}-1)\mu (dx), \quad u \in \mathbb R^d. \end{equation} For a triangular array $(X_{jn})_{1\leq j \leq n}$, the corresponding compound Poisson vectors $(X_{jn}^{\tilde{}})_{1\leq j \leq n}$ are again assumed to have row-wise independent entries.

By $X_n \overset{d}{\sim} Y_n$ we mean that, if either side converges in distribution along a sub-sequence, then so does the other along the same sequence.

We can show the proposition about compound Poisson approximation: Let $(X_{jn})_{1\leq j \leq n}$, $n\in \mathbb N$, be a null array of random vectors in $\mathbb R^d$. Fix $h > 0$, and define: $$b_{jn}= E(X_{jn}\,;\,|X_{jn}|\leq h).$$ Then $$\sum_j X_{jn} \overset{d}{\sim} \sum_j \left\{ (X_{jn} - b_{jn})^{\tilde{}} +b_{jn} \right\}$$

(This the Proposition 7.11 from Foundations of Modern Probability, Third edition)

Question:

Suppose $E[X_{jn}]=0$ for all $j$ and $n$. How to show this?: $$\sum_j X_{jn} \overset{d}{\sim} \sum_j X_{jn}^{\tilde{}} $$ Intuitively, when $h \to \infty$, we have $b_{jn}\to 0$.

I do not know if $E[X_{jn}]=0$ is sufficient. If this condition is not sufficient, I also have that: \begin{equation}\label{I}\tag{I} \sum_{j=1}^n \mathbf{v}(X_{jn})\leq C< \infty,\quad \forall \, n \end{equation} where $X_{jn}=(X_{jn_{1}},..., X_{jn_{d}})$ and: $$\mathbf{v}(X_{jn}):=\hbox{trace} \left(E\left[X_{jn}X_{jn}^{\,\,'}\right]\right)= \sum_{k=1}^d \hbox{var}(X_{jn_{k}})$$ Note that $X_{jn_{k}}$ is unidemsional, $E[X_{jn_{k}}]=0$ and $\hbox{var}(X_{jn_{k}})= E[(X_{jn_{k}})^2]$.

Let $(X_{jn})_{1\leq j \leq n}$, $n\in \mathbb N$, be a triangular array of random vectors in $\mathbb R^d$ (the $X_{jn}$ are understood to be independent in $j$ for fixed $n$.). We say that the triangular array is null if $X_{jn} \overset{p}{\to} 0$, as $n \to \infty$, uniformly in $j$, i.e.,: \begin{equation} \lim_{n \to \infty} \max_{1\leq j\leq n} P( \, |X_{jn}|> \epsilon \, )=0, \quad (\forall \, \epsilon>0). \end{equation} For any random vector $X$ with distribution $\mu$, we introduce an associated compound Poisson random vector $X^{\tilde{}}$ (Note the tilde, please) with characteristic measure $\mu$, i.e.: \begin{equation} \log \varphi_{X^{\tilde{}}}(u) = \int_{\mathbb R^d} (e^{iux}-1)\mu (dx), \quad u \in \mathbb R^d. \end{equation} For a triangular array $(X_{jn})_{1\leq j \leq n}$, the corresponding compound Poisson vectors $(X_{jn}^{\tilde{}})_{1\leq j \leq n}$ are again assumed to have row-wise independent entries.

By $X_n \overset{d}{\sim} Y_n$ we mean that, if either side converges in distribution along a sub-sequence, then so does the other along the same sequence.

We can show the proposition about compound Poisson approximation: Let $(X_{jn})_{1\leq j \leq n}$, $n\in \mathbb N$, be a null array of random vectors in $\mathbb R^d$. Fix $h > 0$, and define: $$b_{jn}= E(X_{jn}\,;\,|X_{jn}|\leq h).$$ Then $$\sum_j X_{jn} \overset{d}{\sim} \sum_j \left\{ (X_{jn} - b_{jn})^{\tilde{}} +b_{jn} \right\}$$

(This the Proposition 7.11 from Foundations of Modern Probability, Third edition)

Question:

Suppose $E[X_{jn}]=0$ for all $j$ and $n$. How to show this?: $$\sum_j X_{jn} \overset{d}{\sim} \sum_j X_{jn}^{\tilde{}} $$ Intuitively, when $h \to \infty$, we have $b_{jn}\to 0$.

I do not know if $E[X_{jn}]=0$ is sufficient. If this condition is not sufficient, I also have that: \begin{equation}\label{I}\tag{I} \sum_{j=1}^n v(X_{jn})\leq C< \infty,\quad \forall \, n \end{equation}

where $v(X_{jn}):=\hbox{trace} \left(E(X_{jn}X_{jn}^{\,\,'})\right)= \sum_{k=1}^d var(X_{jn_{k}})$ and $X_{jn}=(X_{jn_{1}},..., X_{jn_{d}})$. Note that $X_{jn_{k}}$ is unidemsional, $E[X_{jn_{k}}]=0$ and $var(X_{jn_{k}})= E[(X_{jn_{k}})^2]=0$.

My attempt

Note that the compound Poisson random vector $X_{jn}$ is such that $X_{jn}^{\tilde{}}= \sum_{\kappa=1}^N X_{\kappa;jn}$, where $(X_{\kappa;jn})_{\kappa=1}^\infty$ are iid copies of $X_{jn}$ and $N \sim \hbox{Poisson}(1)$. So $$(X_{jn} - b_{jn})^{\tilde{}}= X_{jn}^{\tilde{}} - N b_{jn}$$ Now, agregating we have: $$ \begin{aligned} \sum_j \left\{ (X_{jn} - b_{jn})^{\tilde{}} +b_{jn} \right\}&=\sum_j (X_{jn} - b_{jn})^{\tilde{}} +\sum_j b_{jn}\\ &=\sum_j X_{jn}^{\tilde{}} - (N-1)\sum_j b_{jn} \end{aligned} $$ Thus, it remains to show that $\sum_j b_{jn}=0$ as $h \to \infty$.

For this, I will do $h \in \mathbb N$. Since \begin{equation}\label{II}\tag{II} X^h_{jn}:= X_{jn} \,\mathbf{1}_{[|X|_{jn}< h]} \overset{a.s}{\longrightarrow} X_{jn}, \quad(h \to \infty) \end{equation} my first attempt was to use the Dominated Convergence Theorem for the random vectors for $X^h_{jn}$. But I can't find $Y$ with $E[Y]< \infty$ such that $|X^h_{jn}|\leq |Y|$, for all $h$. So I think that (\ref{I}) is necessary. Using this condition I will try to show uniform integrability and conclude.

First, suppose that $d=1$. Defining $S_h := \sum_{j=1}^n X^h_{jn}$, $h \in \mathbb N$. Remember that $X_{in}$ and $X_{jn}$ are independent. So, using this result, we have that (i) $X_{in}^h$ and $X_{jn}^h$ are independent. Moreover, note that (ii) $|X^h_{jn}|\leq |X_{jn}|$. Thus, we have:
$$E[S_h^2]\overset{(i)}{=} \sum_{j} E[(X^h_{jn})^2] \overset{(ii)}{\leq} \sum_{j} E[X_{jn}^2]\overset{(\ref{I})}{\leq}C$$ This implies that: \begin{equation}\label{III}\tag{III} \sup_{h }E [S_h^2]< \infty \end{equation} By (\ref{II}), we have that $S_h = \sum_{j=1}^n X^h_{jn}\overset{d}{\longrightarrow} \sum_{j=1}^n X_{jn}$, as $h \to \infty$. This together with (\ref{III}), we conclude that $$\sum_{j=1}^n b_{jn}= E \left[S_h \right] \longrightarrow E \left[\sum_{j=1}^n X_{jn}\right]=0, \quad (h \to \infty).$$

Well, I would like your help to know if:

a) For the case $d=1$ I am not making any big mistakes. Is my strategy right?

b) If the case $d=1$ is correct, how can we generalize to higher dimensions? Honestly, I don't know if there is any uniform integrability result using the variance for vectors that I defined above.

Let $(X_{jn})_{1\leq j \leq n}$, $n\in \mathbb N$, be a triangular array of random vectors in $\mathbb R^d$ (the $X_{jn}$ are understood to be independent in $j$ for fixed $n$.). We say that the triangular array is null if $X_{jn} \overset{p}{\to} 0$, as $n \to \infty$, uniformly in $j$, i.e.,: \begin{equation} \lim_{n \to \infty} \max_{1\leq j\leq n} P( \, |X_{jn}|> \epsilon \, )=0, \quad (\forall \, \epsilon>0). \end{equation} For any random vector $X$ with distribution $\mu$, we introduce an associated compound Poisson random vector $X^{\tilde{}}$ (Note the tilde, please) with characteristic measure $\mu$, i.e.: \begin{equation} \log \varphi_{X^{\tilde{}}}(u) = \int_{\mathbb R^d} (e^{iux}-1)\mu (dx), \quad u \in \mathbb R^d. \end{equation} For a triangular array $(X_{jn})_{1\leq j \leq n}$, the corresponding compound Poisson vectors $(X_{jn}^{\tilde{}})_{1\leq j \leq n}$ are again assumed to have row-wise independent entries.

By $X_n \overset{d}{\sim} Y_n$ we mean that, if either side converges in distribution along a sub-sequence, then so does the other along the same sequence.

We can show the proposition about compound Poisson approximation: Let $(X_{jn})_{1\leq j \leq n}$, $n\in \mathbb N$, be a null array of random vectors in $\mathbb R^d$. Fix $h > 0$, and define: $$b_{jn}= E(X_{jn}\,;\,|X_{jn}|\leq h).$$ Then $$\sum_j X_{jn} \overset{d}{\sim} \sum_j \left\{ (X_{jn} - b_{jn})^{\tilde{}} +b_{jn} \right\}$$

(This the Proposition 7.11 from Foundations of Modern Probability, Third edition)

Question:

Suppose $E[X_{jn}]=0$ for all $j$ and $n$. How to show this?: $$\sum_j X_{jn} \overset{d}{\sim} \sum_j X_{jn}^{\tilde{}} $$ Intuitively, when $h \to \infty$, we have $b_{jn}\to 0$.

I do not know if $E[X_{jn}]=0$ is sufficient. If this condition is not sufficient, I also have that: \begin{equation}\label{I}\tag{I} \sum_{j=1}^n v(X_{jn})\leq C< \infty,\quad \forall \, n \end{equation}

where $v(X_{jn}):=\hbox{trace} \left(E(X_{jn}X_{jn}^{\,\,'})\right)= \sum_{k=1}^d var(X_{jn_{k}})$ and $X_{jn}=(X_{jn_{1}},..., X_{jn_{d}})$. Note that $X_{jn_{k}}$ is unidemsional, $E[X_{jn_{k}}]=0$ and $var(X_{jn_{k}})= E[(X_{jn_{k}})^2]$.

My attempt

Note that the compound Poisson random vector $X_{jn}$ is such that $X_{jn}^{\tilde{}}= \sum_{\kappa=1}^N X_{\kappa;jn}$, where $(X_{\kappa;jn})_{\kappa=1}^\infty$ are iid copies of $X_{jn}$ and $N \sim \hbox{Poisson}(1)$. So $$(X_{jn} - b_{jn})^{\tilde{}}= X_{jn}^{\tilde{}} - N b_{jn}$$ Now, agregating we have: $$ \begin{aligned} \sum_j \left\{ (X_{jn} - b_{jn})^{\tilde{}} +b_{jn} \right\}&=\sum_j (X_{jn} - b_{jn})^{\tilde{}} +\sum_j b_{jn}\\ &=\sum_j X_{jn}^{\tilde{}} - (N-1)\sum_j b_{jn} \end{aligned} $$ Thus, it remains to show that $\sum_j b_{jn}=0$ as $h \to \infty$.

For this, I will do $h \in \mathbb N$. Since \begin{equation}\label{II}\tag{II} X^h_{jn}:= X_{jn} \,\mathbf{1}_{[|X|_{jn}< h]} \overset{a.s}{\longrightarrow} X_{jn}, \quad(h \to \infty) \end{equation} my first attempt was to use the Dominated Convergence Theorem for the random vectors for $X^h_{jn}$. But I can't find $Y$ with $E[Y]< \infty$ such that $|X^h_{jn}|\leq |Y|$, for all $h$. So I think that (\ref{I}) is necessary. Using this condition I will try to show uniform integrability and conclude.

First, suppose that $d=1$. Defining $S_h := \sum_{j=1}^n X^h_{jn}$, $h \in \mathbb N$. Remember that $X_{in}$ and $X_{jn}$ are independent. So, using this result, we have that (i) $X_{in}^h$ and $X_{jn}^h$ are independent. Moreover, note that (ii) $|X^h_{jn}|\leq |X_{jn}|$. Thus, we have:
$$E[S_h^2]\overset{(i)}{=} \sum_{j} E[(X^h_{jn})^2] \overset{(ii)}{\leq} \sum_{j} E[X_{jn}^2]\overset{(\ref{I})}{\leq}C$$ This implies that: \begin{equation}\label{III}\tag{III} \sup_{h }E [S_h^2]< \infty \end{equation} By (\ref{II}), we have that $S_h = \sum_{j=1}^n X^h_{jn}\overset{d}{\longrightarrow} \sum_{j=1}^n X_{jn}$, as $h \to \infty$. This together with (\ref{III}), we conclude that $$\sum_{j=1}^n b_{jn}= E \left[S_h \right] \longrightarrow E \left[\sum_{j=1}^n X_{jn}\right]=0, \quad (h \to \infty).$$

Well, I would like your help to know if:

a) For the case $d=1$ I am not making any big mistakes. Is my strategy right?

b) If the case $d=1$ is correct, how can we generalize to higher dimensions? Honestly, I don't know if there is any uniform integrability result using the variance for vectors that I defined above.