Skorohod Representation Theorem
1 Statement
Skorohod’s Representation Theorem shows that a sequence converging in law can be represented by a sequence with the same laws that converges almost surely on a common probability space. This lets us prove statements about convergence in law using pointwise arguments.
Suppose that \(X_{n}\stackrel{\mathcal{D}}{\to}X\) as \(n \to \infty\) with \(F_{n}(x):=\mathbb{P}(X_{n} \leq x)\) and \(F(x):=\mathbb{P}(X\leq x)\) for \(x \in \mathbb{R}\). Then there exists a probability space \((\Omega', \mathcal{F}',\mathbb{P}')\) and random variables \(\{Y_{n}, n\geq 1\}\) and \(Y\) on it such that \[ Y_{n}\stackrel{\mathcal{D}}{=}X_{n}\ \text{ for every } n,\qquad Y\stackrel{\mathcal{D}}{=}X, \] and \(Y_{n}\stackrel{a.s.}{\to}Y\) as \(n \to \infty\).
2 Skorohod’s Construction
Take \(\Omega'=(0,1)\), \(\mathcal{F}'=\mathcal{B}((0,1))\) and \(\mathbb{P}'\) the Lebesgue measure, \(\mathbb{P}'((a,b])=b-a\). For \(\omega \in \Omega'\) define the generalised inverses \[ Y_n(\omega):= \inf\{ x: \omega \leq F_{n}(x)\},\qquad Y(\omega):= \inf\{ x: \omega \leq F(x) \}. \] By right-continuity of distribution functions, \[ \omega \leq F_{n}(x) \iff Y_{n}(\omega)\leq x,\qquad \omega \leq F(x) \iff Y(\omega)\leq x, \] hence \[ \mathbb{P}'(Y_{n}\leq y) =\mathbb{P}'(\omega \leq F_{n}(y)) =F_{n}(y),\qquad \mathbb{P}'(Y\leq y)= F(y). \] So \(Y_n\) has the law of \(X_n\) and \(Y\) the law of \(X\).
Proof. It remains to show \(Y_{n}\to Y\) a.s. Note that \(Y\) and each \(Y_n\) are non-decreasing in \(\omega\). Recall that \(X_n\Rightarrow X\) means \(F_n(x)\to F(x)\) at every continuity point \(x\) of \(F\), and these points are dense in \(\mathbb{R}\). Fix \(\omega\in(0,1)\).
Lower bound. Let \(x<Y(\omega)\) be a continuity point of \(F\). Then \(F(x)<\omega\), so \(F_n(x)<\omega\) for all large \(n\), which means \(Y_n(\omega)>x\). Hence \(\liminf_n Y_n(\omega)\geq x\), and letting \(x\uparrow Y(\omega)\) gives \(\liminf_n Y_n(\omega)\geq Y(\omega)\).
Upper bound. Let \(\omega<\omega'<1\) and let \(y>Y(\omega')\) be a continuity point of \(F\). Then \(F(y)\geq\omega'>\omega\), so \(F_n(y)>\omega\) for all large \(n\), so \(Y_n(\omega)\leq y\). Hence \(\limsup_n Y_n(\omega)\leq y\), and letting \(y\downarrow Y(\omega')\) gives \(\limsup_n Y_n(\omega)\leq Y(\omega')\).
If \(\omega\) is a continuity point of \(Y\), letting \(\omega'\downarrow\omega\) gives \(\limsup_n Y_n(\omega)\leq Y(\omega)\), so \(Y_n(\omega)\to Y(\omega)\). As \(Y\) is monotone it has at most countably many discontinuities, a Lebesgue-null set, so \(Y_n\to Y\) \(\mathbb{P}'\)-almost surely.
3 Application
If \(X_{n}\stackrel{\mathcal{D}}{\to}X\) and \(\varphi\) is bounded and continuous, then \[ \lim_{n\to \infty}\mathbb{E}[\varphi(X_{n})]=\mathbb{E}[\varphi(X)]. \]
Proof. With \(Y_n,Y\) as above, \(\varphi(Y_n)\to\varphi(Y)\) a.s. by continuity, and \(|\varphi(Y_n)|\leq\sup|\varphi|<\infty\). By the Dominated Convergence Theorem, \(\mathbb{E}\varphi(X_n)=\mathbb{E}\varphi(Y_n)\to\mathbb{E}\varphi(Y)=\mathbb{E}\varphi(X)\).