Showing $Y_n\stackrel{p}\to Z$ where $Y_n=B_nZ+(1-B_n)X$

I am reviewing some of my old class notes again, and I came across the following problem. I think I have solved the problem correctly, but I wanted to see what others here thought. Do you think I answered this problem correctly? If not, can you suggest where I have gone wrong? Thank you, in advance for your help.

Suppose $$Z\sim N(0,1)$$, $$X\sim N(1,1)$$, and independent of $$Z$$ and $$X$$, let $$B_n\sim \text{Bernoulli}\left(1-{1\over{n}}\right)$$ for each $$n \ge 1$$. Define: $$Y_{n} = B_{n}Z+(1-B_{n})X$$ Show that $$Y_{n}\overset{p}{\rightarrow}Z$$.

So, I approached this using the law of total probability as follows:

\begin{eqnarray*} P\left(|Y_{n}-Z|\ge\epsilon\right) & = & P\left(|Y_{n}-Z|\ge\epsilon \left|B_{n}=0\right.\right)P\left(B_{n}=0\right)+\\&&P\left(|Y_{n}-Z|\ge\epsilon\left|B_{n}=1\right.\right)P\left(B_{n}=1\right)\\ & = & P\left(|X-Z|\ge\epsilon\right)P\left(B_{n}=0\right)+P\left(|Z-Z|\ge\epsilon\right)P\left(B_{n}=1\right)\\ & = & P\left(|X-Z|\ge\epsilon\right)P\left(B_{n}=0\right)+P\left(0\ge\epsilon\right)P\left(B_{n}=1\right)\\ & = & P\left(|X-Z|\ge\epsilon\right)P\left(B_{n}=0\right)\begin{aligned} &&& \text{(since \epsilon is strictly greater than 0})\end{aligned} \end{eqnarray*}

Now, recall that for any real numbers, $$u$$,$$v\in[0,1]$$, $$uv\le v$$. So since, by definition of probabilities, $$P\left(|X-Z|\ge\epsilon\right)$$, $$P\left(B_{n}=0\right)\in[0,1]$$, we obtain $$\begin{eqnarray*} P\left(|X-Z|\ge\epsilon\right)P\left(B_{n}=0\right) & \le & P\left(B_{n}=0\right)=\frac{1}{n} \end{eqnarray*}$$

To summarize, we have

$$\begin{eqnarray*} P\left(|Y_{n}-Z|\ge\epsilon\right) & \le & \frac{1}{n} \end{eqnarray*}$$

so, by definition of convergence in probability:

$$\begin{eqnarray*} \underset{n\rightarrow\infty}{\lim}\,P\left(|Y_{n}-Z|>\epsilon\right) & \le & \underset{n\rightarrow\infty}{\lim}\frac{1}{n}=0\\&& \end{eqnarray*}$$

so $$Y_{n}\overset{p}{\rightarrow}Z$$. QED.

• You could begin by noting that the event $|Y_n-Z|\gt \epsilon$ must be a subset of the event $B_n=0$ and you'd be done, because you could jump straight to the statement after "to summarize." – whuber May 11 at 20:10
• @whuber, could you explain this a bit more? How can I tell that $|Y_n-Z| \ge \epsilon$ must be a subset of event $B_n=0$? I'm having a tough time wrapping my brain around it for some reason. – StatCurious May 12 at 5:06
• When $B_n=1, Y_n=Z,$ implying $|Y_n-Z|=0 \lt \epsilon.$ The contrapositive of this statement is $|Y_n-Z|\ne 0$ implies $B_n\ne 1.$ Finish up by noting $|Y_n-Z|\gt \epsilon$ is a subset of $|Y_n-Z|\ne 0$ and $B_n\ne 1$ is equivalent to $B_n=0.$ – whuber May 12 at 15:22
• Ah, yes. That makes perfect sense, @whuber. Thank you! – StatCurious May 13 at 20:19