# Convergence by probability and transformation

Question 1:

If $$X_n$$ converges in probability to $$X$$, can I apply the continuous mapping theorem to say that $$\dfrac{X_n}{n}$$ converges in probability to $$\dfrac{X}{n}$$?

Continuous mapping theorem as per wikipedia: Suppose $$g(\cdot)$$ is a continuous function. If $$X_n$$ converges in probability to $$X$$, then $$g(X_n)$$ converges in probability to $$g(X)$$.

Question 2:

My metrics book stated that, if $$X_n$$ converges in probability to $$X$$, then $$\dfrac{nX_n}{n - 3}$$ will converge in probability to $$X$$. Why? Why is $$\dfrac{n}{n - 3}$$ consistent to $$1$$?

• For the sake of clarity, can you please specify exactly what you mean by the "continuous mapping theorem"? Commented May 11, 2021 at 14:22

I'm not sure how you can use it since $$n$$ also goes to infinity, but using the definition is also simple. For $$n\geq 1$$, we have $$\mathbb P(|X_n/n-X/n|>\epsilon)\leq \mathbb P(|X_n-X|>\epsilon)$$ because the LHS term is getting smaller when we divide it by $$n$$. Taking the limit makes the right of the inequality $$0$$, which makes the left of the inequality $$0$$ as well.
For the second one, you can use the theorem. You have $$Z_n=\left[ X_n\ \ Y_n\right]\rightarrow [X\ \ Y]$$ where $$Y_n=n/(n-3)\rightarrow 1 = Y$$, so $$g(Z_n)=X_nY_n\rightarrow XY=X$$.
In your first question, continuous mapping theorem is not applicable because the resulting limit should not depend on $$n$$. Moreover, $$g(\cdot)$$ in CMT is a continuous function that also should not depend on $$n$$, otherwise you would get a sequence of functions.