Convergence of random variables Trying to understand the solution given to this homework problem:
Define random variables $X$ and $Y_n$ where $n=1,2\ldots%$ with probability mass functions:
$$
f_X(x)=\begin{cases} 
\frac{1}{2} &\mbox{if } x = -1 \\ 
\frac{1}{2} &\mbox{if } x = 1  \\
0 &\mbox{otherwise}
\end{cases} and\; f_{Y_n}(y)=\begin{cases} 
\frac{1}{2}-\frac{1}{n+1} &\mbox{if } y = -1 \\ 
\frac{1}{2}+\frac{1}{n+1} &\mbox{if } y = 1  \\
0 &\mbox{otherwise}
\end{cases}
$$
Need to show whether $Y_n$ converges to $X$ in probability.
From this I can define the probability space $\Omega=([0,1],U)$ and express the random variables as functions of indicator variables as such:
$X = 1_{\omega > \frac{1}{2}} - 1_{\omega < \frac{1}{2}}$
and
$Y_n = 1_{\omega < \frac{1}{2}+\frac{1}{n+1}} - 1_{\omega > \frac{1}{2}+\frac{1}{n+1}}$
And from the definition of convergence in probability, we need find to show that
$P\{|Y_n-X|>\epsilon\}$ does or does not converge to zero. Which can be written as:
$P\{|1_{\omega < \frac{1}{2}+\frac{1}{n+1}} - 1_{\omega > \frac{1}{2}+\frac{1}{n+1}} - 1_{\omega > \frac{1}{2}} + 1_{\omega < \frac{1}{2}}| > \epsilon \}\;\;(1)$
Now it's easy to see that $\epsilon < 2$ for this to hold, but the solution given states that:
$P\{|Y_n-X|>\epsilon\} = 1 - \frac{1}{n+1} \;\; (2)$
Thus $Y_n$ does not converge in probability to $X$.
My problem is that I don't see the reasoning between (1) and (2). Can anyone shed some insight into intermediate steps/reasoning required to make this step?
 A: You're told that
$$
  P(X=1)=P(X=-1)=1/2 \, ,
$$
and
$$
  P(Y_n=1)=\frac{1}{2} + \frac{1}{n+1}  \;\;\;, \qquad P(Y_n=-1)=\frac{1}{2} - \frac{1}{n+1} \;\;\;,
$$
for $n\geq 1$, and you're asked whether or not $Y_n$ converges to $X$ in probability, which means that
$$
  \lim_{n\to\infty} P(|Y_n-X|\geq \epsilon) = 0 \, , \qquad (*)
$$
for every $\epsilon>0$.
I will assume that $X$ is independent of the $Y_n$'s.
It is not the case that $Y_n$ converges in probability to $X$, because $(*)$ does not hold for every $\epsilon>0$. 
For instance, if we take $\epsilon=1$, then
$$
  P(|Y_n-X|\geq 1)=P(Y_n=1, X=-1) + P(Y_n=-1,X=1) 
$$
$$
  = P(Y_n=1)P(X=-1) + P(Y_n=-1)P(X=1) 
$$
$$
  = \left(\frac{1}{2} + \frac{1}{n+1}\right) \cdot \frac{1}{2} + \left(\frac{1}{2} - \frac{1}{n+1}\right) \cdot \frac{1}{2} = \frac{1}{2} \, ,
$$
for every $n\geq 1$.
A: Given random variables $X, Y_1, Y_2, \ldots, $ with probability mass functions
$$p_X(x)=\begin{cases} 
\frac{1}{2}, &\text{if}~ x = -1, \\ 
\frac{1}{2}, &\text{if}~ x = +1  \\
0 &\text{otherwise,}
\end{cases} ~~~~~\text{and}~~~ p_{Y_n}(y)=\begin{cases} 
\frac{1}{2}-\frac{1}{n+1}, &\text{if}~ y = -1, \\ 
\frac{1}{2}+\frac{1}{n+1}, &\text{if}~ y = +1,  \\
0 &\text{otherwise,}
\end{cases}$$
it is straightforward to show that the sequence $\{Y_n\} = (Y_1, Y_2, \ldots)$, 
of random variables converges in distribution to $X$ (cf. 
Xi'an's comment on the question). However, the
question of whether $\{Y_n\}$ converges in probability to $X$ depends on
our assumptions about the joint probability mass functions
$p_{X,Y_n}(x,y)$.


*

*If we assume, as Zen does, that $X$ is independent of each
$Y_n$, then, as Zen shows, $\{Y_n\}$ does not converge in probability
to $X$. This is not due to the special circumstance of these random
variables being kissing cousins of Bernoulli random variables but holds
more generally.  If $X$ is nondegenerate random variable, that
is $X$ does not have constant value with probability $1$, then no sequence 
$\{Y_n\}$ of random variables independent of $X$ can converge in
probability to $X$.  Note that if $X$ is a discrete nondegenerate
random variable that is independent of a discrete random variable $Y_n$
then
$$P\{Y_n = X\} = \sum_i p_X(u_i)p_{Y_n}(u_i) 
\leq \sqrt{\sum_i \left(p_X(u_i)\right)^2
\sum_i \left(p_{Y_n}(u_i)\right)^2}$$
Now, $\sum_i \left(p_{Y_n}(u_i)\right)^2$
is just $P\{A = B\}$ where $A$ and $B$ are independent identically
distributed discrete random variables with common pmf
$p_{Y_n}(\cdot)$, and thus the sum is at most $1$
for all values of $n$.  On the other hand, since $X$ is 
nondegenerate by assumption so that $p_X(u_i) < 1$ for all $i$
while $\sum_i p_X(u_i) = 1$,
it follows that $\sum_i \left(p_X(u_i)\right)^2$
cannot equal $1$;
it is a constant that is strictly
smaller than $1$, say $1-\delta$ for some
$\delta > 0$. Hence
$$\begin{align}
P\{Y_n = X\} &\leq \sqrt{\sum_i \left(p_X(u_i)\right)^2
\sum_i \left(p_{Y_n}(u_i)\right)^2}\\
&\leq \sqrt{1-\delta}\cdot 1\\
&\Rightarrow~~
\lim_{n\to\infty}P\{Y_n = X\} < 1.
\end{align}$$ 
For continuous random variable $X$ independent of
continuous random variable $Y_n$, $P\{Y_n = X\} = 0$.
Thus, we have the following result:



If $X$ is a nondegenerate random variable, then
  no sequence $\{Y_n\}$ of random variables, each of
  which is independent of $X$, can converge in probability 
  to $X$.



*

*For $n \geq 3$, if $X$ and $Y_n$ are assumed to be 
dependent random variables with joint probability mass function 
$$\begin{alignat}{4}
&p_{X,Y_n}(-1, +1)~& &= \frac{2}{n+1},& \qquad
&p_{X,Y_n}(+1, +1)~& &= \frac{1}{2}- \frac{1}{n+1},&\\
&p_{X,Y_n}(-1, -1)~& &= \frac{1}{2} - \frac{2}{n+1},& \qquad
&p_{X,Y_n}(+1, -1)~& &= \frac{1}{n+1},&\\
\end{alignat}$$
then it is easy to verify that the marginal probability
mass functions are as specified. Also,
$$P\{Y_n \neq X\} = \frac{3}{n+1} \to 0 ~ \text{as}~ n \to \infty$$
and so the sequence $\{Y_n\}$ converges in probability to $X$.

*For $n \geq 3$, if $X$ and $Y_n$ are assumed to be 
dependent random variables with joint probability mass function 
$$\begin{alignat}{4}
&p_{X,Y_n}(-1, +1)~& &= \frac{1}{2}- \frac{1}{n+1},& \qquad
&p_{X,Y_n}(+1, +1)~& &= \frac{2}{n+1},&\\
&p_{X,Y_n}(-1, -1)~& &=  \frac{1}{n+1},& \qquad
&p_{X,Y_n}(+1, -1)~& &= \frac{1}{2} -\frac{2}{n+1},&\\
\end{alignat}$$
then it is easy to verify that the marginal probability
mass functions are as specified. Also,
$$P\{Y_n \neq X\} = 1 - \frac{3}{n+1} \to 1 ~ \text{as}~ n \to \infty$$
and so the sequence $\{Y_n\}$ does not converge in probability to $X$:
it does, however, converge in probability to $-X$.

*If we assume the joint probability mass function alternates
(according as $n$ is odd or even) between the two joint
mass functions in the above two bullet point, then we get
that $\{Y_n\}$ does not converge in probability at all.

Note to OP:  The question you have posed "Does $\{Y_n\}$
converge in probability to $X$?" can be made to have whatever
answer you like by choosing the joint distribution of $X$ and $Y_n$
appropriately. 


*

*You can choose each $Y_n$ to be independent
of $X$ in which case convergence in probability cannot occur.
This is proved in Zen's answer to your question
for your particular random variables,
and in more generality above.

*You can choose the joint distributions so that $\{Y_n\}$
converges in probability to $X$, as described above.  Your own
partial answer can be modified to ensure that $\{Y_n\}$
converges to $X$ in probability because the sequence has
the far stronger property of converging almost surely.

*You can choose the joint distributions so that $\{Y_n\}$
converges in probability but converges to $-X$, not to $X$, 
as described above. 

*You can choose the joint distributions so that convergence
in probability does not occur at all, whether to the specified
$X$ or to any other random variable.
