Let $\mathbb P(X=1) = \mathbb P(X=-1) = 1/2$. Define

$$X_n = \begin {cases} X & \text{with probability } 1- \frac{1}{n}\\ e^n & \text{with probability } \frac{1}{n} \end {cases}$$

Does $X_n \leadsto X$ (i.e., converge in distribution)? Does $X_n \stackrel{P}{\rightarrow} X$? Does $\mathbb{E}[(X-X_n)^2] \to 0$?

I think $X_n \leadsto X$ because just by picturing the CDF as n goes to infinity, their CDFs converge. Can I get some confirmation on this?

But for convergence in probability, I'm having trouble with the limits.

To do this, what I'm trying to do is find $\mathbb P(|X_n-X| \leq \epsilon)$ and then just take 1 minus that as shown on this site: http://www.statlect.com/prbcon1.htm

If $X_n = X$, then $|X_n - X|$ is just $0$ so for all $n$, $|X_n -X| < \epsilon$ But if $X_n = e^n$ then I have the two cases for $X$. For $X=1$, $|X_n-X| = |e^n-1| \leq \epsilon$. and For $X=-1$, $|X_n-X| = |e^n +1| = e^n + 1 \leq \epsilon$. This is where I am stuck. I'm trying to find conditions relating $n$ and $\epsilon$ similiar to here the example on the site. But I'm not sure how to get that from $|e^n - 1| \leq \epsilon$ and $e^n + 1 \leq \epsilon$ .

Am I even on the right track?

  • 4
    $\begingroup$ You are interpreting $X$ and $X_n$ as deterministic quantities, not as random variables. So the sentence "If $X_n=X$ then $|X-X_n|=0$ for all $n$'s" does not make sense. Those are random variables, e.g. $|X-X_n|=0$ with probability $1-(1/n)$ and $e^n$ with probability $1/n$. From there, you can conclude about convergence in probability and $L_2$ [lack of] convergence. $\endgroup$
    – Xi'an
    Commented Feb 23, 2012 at 8:15

1 Answer 1


To prove that $X_n \stackrel{P}{\rightarrow} X$ you have to show that, for any given $\epsilon>0$, the sequence of real numbers $\{p_n\}_{n\geq 1}$ defined by $$ p_n = P\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon\} $$ has limit $0$. What happens at the ``beginning'' of the sequence does not change its limit. For $n\geq \log(1+\epsilon)$, we have $$ P\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon\} = P\{\omega:X_n(\omega)=e^n\}=\frac{1}{n} \, . $$ Hence, $p_n\to 0$, and therefore $X_n \stackrel{P}{\rightarrow} X$.

Since convergence in probability implies convergence in distribution, we also have $X_n \leadsto X$.

Now, since $$ \begin{align*} \mathbb{E}[(X_n-X)^2]&=0^2\times\left(1-\frac{1}{n}\right) + (e^n+1)^2\times\frac{1}{n}\times\frac{1}{2}+(e^n-1)^2\times\frac{1}{n}\times\frac{1}{2}\\ & = \frac{(e^{2n}+1)}{n}\to\infty \, , \end{align*} $$ we don't have convergence in quadratic mean.

  • 5
    $\begingroup$ (+1) Welcome to the site. A couple notes: (a) We generally do not give full answers to homework problems on this site, except in exceptional circumstances . Providing hints as answers is strongly preferred. (FAQ). (b) The last part can be simplified a little by noting that immediately from the definition of $X_n$ we have $\mathbb E((X_n - X)^2) \geq \frac{1}{n} (e^n-1)^2 \to \infty$. $\endgroup$
    – cardinal
    Commented Feb 25, 2012 at 16:39
  • $\begingroup$ Maybe a $2$ is missing in the denominator? $$\mathbb E((X_n - X)^2) \geq \frac{1}{2n} (e^n-1)^2 \to \infty$$ $\endgroup$
    – Zen
    Commented Feb 25, 2012 at 20:40
  • $\begingroup$ No problem. newer users often aren't aware of some of the finer details of what happens on the site or how. So, I just wanted to point you to a useful resource in case you hadn't seen it. Cheers. $\endgroup$
    – cardinal
    Commented Feb 25, 2012 at 20:42
  • $\begingroup$ No, it is not necessary to have a 2 in the denominator. Think of the range of values that $X$ can take on. $\endgroup$
    – cardinal
    Commented Feb 26, 2012 at 2:41

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