# Product of random variables convergence in probability

I want to prove that if $X_n\overset{P}\to X$ and $Y_n\overset{P}\to Y$, then

$$X_n+Y_n\overset{P}\to X+Y,$$ $$X_nY_n\overset{P}\to XY$$

Proof for the first one I simply use triangle inequality $$\mathbf{P}\left(\omega \mid |(X_n(\omega)+Y_n(\omega))-(X(\omega)+Y(\omega))|<\epsilon\right) \leqslant \mathbf{P}\left(\omega \mid |(X_n(\omega)-X(\omega))|+|Y_n(\omega)-Y(\omega)|<\epsilon\right)\leqslant$$ $$\leqslant \mathbf{P}\left(\omega | |X_n(\omega)-X(\omega)|<\frac{\epsilon}2\right)+\mathbf{P}\left(\omega | |Y_n(\omega)-Y(\omega)|<\frac{\epsilon}2\right)\underset{n}\to 0$$

I could use some ideas to prove convergence for the product.

Obviously,

$$|X_n Y_n-XY| \leq |Y| \cdot |X_n-X|+ |X_n| \cdot |Y_n-Y|.$$

By the triangle inequality,

$$\mathbb{P}(|X_n Y_n-XY| \geq \epsilon) \leq \mathbb{P}(|Y| \cdot |X_n-X| \geq \epsilon/2) + \mathbb{P}(|X_n| \cdot |Y_n-Y| \geq \epsilon/2).$$

We estimate the terms separately. For the first one, note that

$$\mathbb{P}(|Y| \cdot |X_n-X| \geq \epsilon/2)\leq \mathbb{P}(|X_n-X| \geq \epsilon/(2M)) + \mathbb{P}(|Y| \geq M)$$

for any $M>0$. Letting $n \to \infty$, we find

$$\limsup_{n \to \infty} \mathbb{P}(|Y| \cdot |X_n-X| \geq \epsilon/2) \leq \mathbb{P}(|Y| \geq M).$$

Since this holds for any $M$, we can let $M \to \infty$ and, e.g. by the dominated convergence theorem, it follows that the limit $\lim_{n \to \infty} \mathbb{P}(|Y| \cdot |X_n-X| \geq \epsilon/2)$ exists and equals $0$.

The second term is a little bit more difficult, but the idea is the same. Again, we estimate

$$\mathbb{P}(|X_n| \cdot |Y_n-Y| \geq \epsilon/2) \leq \mathbb{P}(|Y_n-Y| \geq \epsilon/(2M)) + \mathbb{P}(|X_n| \geq M). \tag{1}$$

Now, by the triangle inequality,

$$\mathbb{P}(|X_n| \geq M) \leq \mathbb{P}(|X_n-X| \geq M/2) + \mathbb{P}(|X| \geq M/2). \tag{2}$$

Plugging $(2)$ into $(1)$ yields

$$\mathbb{P}(|X_n| \cdot |Y_n-Y| \geq \epsilon/2) \leq \mathbb{P}(|Y_n-Y| \geq \epsilon/(2M))+ \mathbb{P}(|X_n-X| \geq M/2) + \mathbb{P}(|X| \geq M/2).$$

The first two terms on the right-hand side converge to $0$ as $n \to \infty$ because $X_n \to X$ and $Y_n \to Y$ in probability. After letting $n \to \infty$ we also let $M \to \infty$ and this proves

$$\lim_{n \to \infty} \mathbb{P}(|X_n| \cdot |Y_n-Y| \geq \epsilon/2) =0.$$

• Could you please explain exactly how you used the dominated convergence theorem in this case to show that the limit is in fact 0? – wrb98 Aug 4 '17 at 23:38
• @Will $u_M(\omega) := 1_{\{|Y(\omega)| \geq M\}}$, then $|u_M| \leq 1 \in L^1(\mathbb{P})$ and $u_M(\omega) \to 0$ as $M \to \infty$: Hence, by domainted convergence, $$\mathbb{P}(|Y| \geq M) = \int_{\Omega} u_M(\omega) \, d\mathbb{P}(\omega) \xrightarrow[]{M \to \infty} 0.$$ – saz Aug 5 '17 at 4:36