The continuous mapping theorem states that

Let $g: R^n \rightarrow R^k $ be continuous in every point of a set $C$ such that $\mathbb P\left(X\in C\right)=1$.

If $X_n \xrightarrow{d} X $ then ${g(X_{n})\stackrel{d}{\rightarrow}g(X).}$

But I have a counterexample for this theorem

Let $X_n$ is a sequence of two dimension random variable $X_n =(X_n^1,X_n^2)$ such as $X_n^1 = N(0,1)$ and $X_n^2 = -X_n^1$. So we have $$X_n =(X_n^1,X_n^2)\xrightarrow{d} (N(0,1),N(0,1)) $$ Given $g: R^2\rightarrow R $ and $g(x,y) = x+y$. $g$ is so a continuous function in $R^2$. Arcording to the continuous mapping theorem, we must have $${g(X_{n}) = g(X_n^1,X_n^2)\stackrel{d}{\rightarrow}g(N(0,1),N(0,1)) = N(0,1)+N(0,1) = N(0,\sqrt{2})}$$

But $$g(X_{n}) = g(X_n^1,X_n^2) = X_n^1 + X_n^2 = X_n^1 -X_n^1 = 0 $$

What is the error in my arguments?

Thank you in advance.

  • $\begingroup$ Your error is that $$X_n \not\xrightarrow{d} (N(0,1), N(0,1))$$ $\endgroup$ – Tryss Jul 3 '15 at 13:24
  • $\begingroup$ What is $N(0,1)$ in your question? A random variable or a distribution? It is important to keep those concepts carefully apart. $\endgroup$ – drhab Jul 3 '15 at 13:30
  • $\begingroup$ $X_n^1 = N(0,1)$ means $X_n^1 = X$ which $X$ is a standard normal random variable. $\endgroup$ – NN2 Jul 3 '15 at 13:35
  • $\begingroup$ It seems that you abusively assume that the first and second component of the limiting pair are independent. They are not. $\endgroup$ – drhab Jul 3 '15 at 13:49

If every $X_n^1$ has standard normal distribution and $X_n^2=-X_n^1$ then: $$X_n=(X_n^1,X_n^2)\stackrel{d}{\rightarrow}(U,V)$$ where $(U,V)$ has a bivariate normal distribution such that $U$ and $V$ both have standard normal distribution and $U+V=0$.

So we have: $$g(X_n^1,X_n^2)=0=g(U,V)$$ for each $n$.

  • $\begingroup$ But $U$ and $V$ are not necessarily dependant. We note $U$ is a standard normal distribution variable and $Z$ is another standard normal distribution variable. $U$ and $Z$ are independant. We have $X_n^1 = U \Rightarrow$ $X_n^1 \xrightarrow{d} U$ and $X_n^2 = -U$, so $\forall {a}, P(X_n^2 <a) = P(-U<a) = P(Z<a) \xrightarrow{n \infty} P(Z<a)$. By the definition of convergence in distribution, we have $$X_n^2 \xrightarrow{d} Z$$ So, $$(X_n^1,X_n^2) \xrightarrow{d} (U,Z)$$ which $U$ and $Z$ are independant. $\endgroup$ – NN2 Jul 3 '15 at 14:00
  • $\begingroup$ It is not legal to conclude $\left(X_{n}^{1},X_{n}^{2}\right)\stackrel{d}{\rightarrow}\left(U,V\right)$ purely on base of $X_{n}^{1}\stackrel{d}{\rightarrow}U\wedge X_{n}^{2}\stackrel{d}{\rightarrow}V$. $\endgroup$ – drhab Jul 3 '15 at 14:06
  • $\begingroup$ I think you are right. Do you know (or do you know how to find) which conditions we must have in order to conclude $(X_n^1,X_n^2)->(U,V)$ ? $\endgroup$ – NN2 Jul 3 '15 at 14:10
  • $\begingroup$ Have a look here, and expecially look for convergence in distribution of random vectors. $\endgroup$ – drhab Jul 3 '15 at 14:16

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