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.

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    $\begingroup$ 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." $\endgroup$ – whuber May 11 at 20:10
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    $\begingroup$ @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. $\endgroup$ – StatCurious May 12 at 5:06
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    $\begingroup$ 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 $B_n=0.$ $\endgroup$ – whuber May 12 at 15:22
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    $\begingroup$ Ah, yes. That makes perfect sense, @whuber. Thank you! $\endgroup$ – StatCurious May 13 at 20:19

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