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If we have a sequence of variables $X_n$ that converges in distribution to $X$, and a sequence $Y_n$ that converges in distribution to $Y$, then does $X_nY_n$ converge in distribution to $XY$. Assume $X_n$ and $Y_n$ are independent.

Weirdly, I can’t find a theorem on this in my textbook on asymptotics.

user56834
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    Hint: if $X_nY_n$ converges in distribution, so does $\log |X_nY_n|$. – Xi'an May 28 '18 at 12:43
  • @Xi'an Thanks! This makes me think of a follow up question: This argument does not work to say that for any arbitrary function$f$ we have that $f(X_n,Y_n)$ converges in distribution to $f(X,Y)$. Is that nevertheless the case? – user56834 May 28 '18 at 14:59

1 Answers1

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Yes, $X_n\cdot Y_n$ converges in distribution to XY.

Method 1 :

$X_n$ $\rightarrow$ $X$

$\implies$ $\lim_{n\rightarrow ∞}$ ⁡$F(X_n)$ = $F(X)$

Here, $F(X_n)$ and $F(X)$ are the cumulative distribution functions of $X_n$ and $X$, respectively.

$Y_n$ $\rightarrow$ $Y$

$\implies$ $\lim_{n\rightarrow ∞}$ ⁡$F(Y_n)$ = $F(Y)$

Again, $F(Y_n)$ and $F(Y)$ are the cumulative distribution functions of $Y_n$ and $Y$, respectively.

$\lim_{n\rightarrow ∞}$ $F(X_n\cdot Y_n)$ = $\lim_{n\rightarrow ∞}$ $F(X_n)\cdot \lim_{n\rightarrow ∞}$ $F(Y_n)$ [Due to independence]

$\implies$ $\lim_{n\rightarrow ∞}$ $F(X_n\cdot Y_n)$ = $F(X)\cdot F(Y)$

$\implies$ $X_n\cdot Y_n$ $\rightarrow$ $XY$

Method 2 :

Check out the link provided here. https://en.wikipedia.org/wiki/Proofs_of_convergence_of_random_variables

We know, convergence in probability implies convergence in distribution but the opposite is not true. Hence, if convergence of two sequences in probability implies joint convergence in probability, it will surely converge into distribution assuming the sequence of random variables are independent.

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ImtiazX
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    You seem to intend "$F(A)$" to mean the CDF of a random variable $A$. For instance, if $A$ represents the outcome of a fair coin with values $0$ and $3$, then $F(A)$ is a locally constant function with jumps of $1/2$ at $0$ and $3.$ If, then, all $X_n$ and $Y_n$ are equal to $A$, then $F(X_nY_n)$ has a jump of $3/4$ at $0$ and a jump of $1/4$ at $9$--but obviously that does not equal the product of $F(X_n)$ and $F(Y_n),$ nor does it equal the limit of those products. Either your notation needs cleaning up or your argument is incorrect. – whuber May 28 '18 at 18:47
  • Your answer has a basic misconception: The distribution function of the rv $Z_n = X_nY_n$ is not the product of the marginal distribution functions. This factoring result holds for the joint distribution when the variables are independent, but not for the distribution of their product, independent or not. – Alecos Papadopoulos Jun 21 '18 at 21:54
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    This proof has serious notational problems. – zhaofeng-shu33 Mar 23 '21 at 02:14