Consider two [[Random Variable|random variables]] $T$ and $U$, that both converge either [[Modes of Convergence#Convergence Almost Surely|almost surely]] or [[Modes of Convergence#Convergence in Probability|in probability]]. We can combine their limits using addition or multiplication or division. Assume: $ T_n \xrightarrow[n\to \infty]{a.s./ \, \mathbf P}T \qquad \text{and} \qquad U_n \xrightarrow[n\to \infty]{a.s./ \, \mathbf P}U \qquad $ Then: $ \begin{align} T_n+U_n &\xrightarrow[n\to \infty]{a.s./ \, \mathbf P}T+U \\[8pt] T_n*U_n &\xrightarrow[n\to \infty]{a.s./ \, \mathbf P}T_U \\[8pt] \frac{T_n}{U_n} &\xrightarrow[n\to \infty]{a.s./ \, \mathbf P} \frac{T}{U} \quad \text{if }U \ne 0 \end{align} $ ## Slutsky’s Theorem For convergence in distribution, the above rules do not generally apply. However, Slutsky's Theorem provides specific cases where combinations are valid if one random variable converges *in probability to a constant*. Assume: $ T_n \xrightarrow[n\to \infty]{(d)}T \qquad \text{and} \qquad U_n \xrightarrow[n\to \infty]{\, \mathbf P}u \qquad $ Then: $ \begin{align} T_n+U_n &\xrightarrow[n\to \infty]{(d)}T+u \\[8pt] T_n*U_n &\xrightarrow[n\to \infty]{(d)}T_u \\[8pt] \frac{T_n}{U_n} &\xrightarrow[n\to \infty]{(d)} \frac{T}{u} \quad \text{if }u\ne 0 \end{align} $ Slutsky’s Theorem holds because: - $T$ converges to a [[Gaussian Distribution]] by [[Central Limit Theorem]] - $U$ converges to some scalar value $u$. - Combining the two, is just an [[Gaussian Distribution#^e84e63|affine transformation]], which does not change the shape of the Gaussian. ## Continuous Mapping Theorem This theorem states that if a r.v. $T_n$​ converges to $T$ (in any of the three modes of convergence), and $f$ is a continuous function, then the transformed random variable $f(T_n)$ also converges to $f(T)$ in the same mode of convergence. $ T_n \xrightarrow [n \to \infty]{a.s./ \, \mathbf P/(d)} T \implies f(T_ n) \xrightarrow [n\to \infty ]{a.s./ \, \mathbf P/(d)}f(T) $