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I have a basic definition of the Monotone Convergence Theorem:

If $\{X_n\}_{n\geqslant 1}$ is a sequence of random variables such that $X_n\nearrow X$ pointwise and $X_n\geqslant 0\ \forall n$, then $E(X_n)\nearrow E(X)$.

My question , how much liberality can made on the convergence part? I mean: which of these (namely - almost sure convergence , convergence in probability, $\&$ convergence in distribution) can be placed in place of the pointwise convergence and still the definition remains exactly same?

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It is also true under the convergence in distribution if we replace the assumption of non-decresingness by stochastic dominance, that is, $\left(\mathbb P\left\{X_n\gt t\right\}\right)_{n\geqslant 1}$ is non-increasing for any $t$. Indeed, we have $$\mathbb E\left[X_n\right]=\int_0^{+\infty}\Pr\left\{X_n\gt t\right\}\mathrm dt, $$ $\Pr\left\{X_n\gt t\right\}\to \Pr\left\{X\gt t\right\}$ for $\lambda$-almost every $t$ (by convergence in distribution) and $\Pr\left\{X_n\gt t\right\}\uparrow \Pr\left\{X\gt t\right\}$. The result thus follows by an application of the monotone convergence theorem, which works when the pointwise convergence is replaced by the almost sure convergence.

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  • $\begingroup$ You mean if $X_n $ increasingly converges **in distribution** to $X$, then $E(X_n)\nearrow E(X)$ ? $\endgroup$
    – Qwerty
    Nov 3, 2016 at 11:11
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    $\begingroup$ @Qwerty Yes, this is what is mean. $\endgroup$ Nov 4, 2016 at 21:19
  • $\begingroup$ To be frank, can you please provide me with an example of a sequence $X_n$ increasingly converging to $X$ in distribution? My head has just gone blank right now! I need to do certain things to convince myself.. $\endgroup$
    – Qwerty
    Nov 4, 2016 at 21:34

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