This is from Hogg and McKean's "Introduction to Mathematical Statistics". At the stage this exercise (number $5.1.1$) is given, he has not yet spoken about convergence in distribution etc. He has only introduced the definition of convergence in probability. We have results from analysis, Chebyshev's theorem, and the continuity of probability theorem at our disposal to prove the following.

Problem: Let $\{a_n\}$ be a sequence of real numbers. Hence, we can also say that $\{a_n\}$ is a sequence of constant (degenerate) random variables. Let $a$ be a real number. Show that $a_n \to a$ is equivalent to $a_n \xrightarrow{P} a$.

My attempt: Let $\{X_n\}$ be a sequence of degenerate random variables such that $X_n = a_n$, and let $X$ be a degenerate random variable such that $X = a$. In one direction, if $a_n \to a$, then for any given $\epsilon>0$, we can find $N$ such that $n > N$ would imply $a_n \in (a-\epsilon,a+\epsilon)$.

Problematic Statement 1: Since $X_n = a_n$ and $X = a$, we have $\mathbb{P}(|X_n-X| \geq \epsilon) = 0$ if $n>N$.

Since we can find such $N$ for any given $\epsilon > 0$, we have that $$\lim_{n\rightarrow\infty} \mathbb{P}(|X_n-X| \geq \epsilon) = 0.$$ Hence $X_n \xrightarrow{P} X$.

In the opposite direction, we are given tha $X_n \xrightarrow{P} X$ which means we are given that $\lim_{n\rightarrow\infty} \mathbb{P}(|X_n-X| \geq \epsilon) = 0.$ The statement $|X_n-X| \geq \epsilon$ is same as $X_n \geq X + \epsilon$ or $X_n \leq X - \epsilon$ which are two disjoint events. So we have that for any given $\delta > 0$, we can find $N$ such that $n > N$ would imply $\mathbb{P}(X_n \geq X + \epsilon)+\mathbb{P}(X_n \leq X - \epsilon) \in [0,\delta)$ (lower limit is zero as probabilities are non-negative). We can find such $N$ for any arbitrarily small $\delta$. As these are degenerate random variables, the probabilities are either zero or one.

Problematic Statement 2: Hence we conclude that $\mathbb{P}(X_n \geq X + \epsilon) = 0$ and $\mathbb{P}(X_n \leq X - \epsilon) = 0$ which in turn implies that $|a_n-a| < \epsilon$.

Other than the two statements I have marked as problematic, there may be more errors. I feel like this proof erroneous let alone be an air tight one.

Please provide feedback as to how I can correct this proof and make it air tight.

Edit: Adding comments to fill the (why?) parts of the answer.

For (why ?) #1. Let $Z_n = |X - X_n|$ (from measure theory, we know $Z_n$ is a random variable but is an exercise in and of itself). Then $$\begin{align}\mathbb{P}(Z_n=|a_n-a|) &= \mathbb{P}(Z_n=|a_n-a||X=a) \times \mathbb{P}(X=a) + \mathbb{P}(Z_n=|a_n-a||X \neq a) \times \mathbb{P}(X \neq a) \\ &= (\mathbb{P}(X_n=a_n)+\mathbb{P}(X_n=2a-a_n)) \times 1 + 0 = 1. \end{align}$$ If $a=a_n$ by some coincidence, the second probability in the last part of the equation won't exist. With this, $Z_n=|a_n-a|$ almost surely but I am not sure if this is same as being degenerate. If we define a degenerate random variable as one which takes a constant value with a probability of $1$, then the above proof has a chance of survival.

For (why ?) #2. With $Z_n$ being degenerate, it maps every $\omega$ to $|a_n-a|$. Either $|a_n-a| \in B$ or not. In case it does, then $\{w\in\Omega:Z(\omega)\in B\} = \Omega$ and is $\emptyset$ otherwise. In other words, no event exists that gets mapped by $Z_n$ to an element of that Borel set $B$ unless $|a_n-a| \in B$.

For (why ?) #3. Now $|a_n-a|$ can be made really small by choosing large enough $n$. So if $|a_n-a| < \epsilon$, and from #2 above, we know that all events are mapped to $|a_n-a|$ by $Z_n$, there won't be any $\omega \in \Omega$ which would get mapped to a number $\geq \epsilon$ as $a_n-a < \epsilon$. Hence that set $\{\omega\in\Omega:Z_n(\omega) \geq \epsilon\}=\emptyset$.

  • $\begingroup$ For (#2) the argument is fine, for (#3) the argument is fine too, although my idea was to say : since $P(Z_n >\epsilon)<1$, and from (#2) we know that $P(Z_n >\epsilon)\in\{0,1\}$, it follows that $P(Z_n >\epsilon) = 0 $. Furthermore from (#2) again $\{Z_n >\epsilon\} \in \{\emptyset,\Omega\}$, but we saw that its probability is zero so necessarily $\{Z_n >\epsilon\} =\emptyset$. $\endgroup$ Commented Oct 20, 2023 at 10:27
  • $\begingroup$ For (#1) however you're making it way too complicated. Here you need to show that $Z_n$ is degenerate in the sense that $Z_n(\omega) = c$ for some constant $c$ and all $\omega\in\Omega$. This should be very direct $\endgroup$ Commented Oct 20, 2023 at 10:29
  • $\begingroup$ Thanks again for the commenting on my answers to the gaps in your answer. For #1, I did think that after all, random variables are functions mapping events to set of reals so we could simply take the difference also as being a function which directly shows that the difference would be degenerate as well. In order to reduce the verbosity of an already verbose post of mine, I avoided mentioning this line of thinking. Thanks once again for your help. Greatly appreciate it! $\endgroup$ Commented Oct 21, 2023 at 5:08

1 Answer 1


Your proof is overall correct, but I guess I would word it a bit differently and provide more justification for some steps.

For the "problematic statement 1" part : although it is always true that $X_n=a_n$ and $X=a$, (i.e. for all $\omega\in\Omega$, we have $X_n(\omega)=a_n$ and $X(\omega)=a$), in probabilistic reasoning what we would rather say is that these events happen almost surely, or equivalently that the events $A_n := \{ X_n=a_n\text{ and }X=a\}$ have probability $\mathbb P(A_n)=1$. With that information we can write

$$\begin{align}\mathbb{P}(|X_n-X| \geq \epsilon) &= \mathbb{P}\big(|a_n-a| \geq \epsilon,A_n\big) \\ &+\mathbb{P}\big(|X_n-X| \geq \epsilon,A_n^c\big)\\ &=\mathbb{P}\big(|a_n-a| \geq \epsilon,A_n\big) +0\\ &\le\mathbb{P}\big(|a_n-a| \geq \epsilon\big) \end{align} $$ And by your argument we know that $\mathbb{P}\big(|a_n-a| \geq \epsilon\big)$ is equal to $0$ for all sufficiently large $n$.

For the "problematic statement 2" part : your argument is correct "in spirit", but a bit incomplete I would say, as you don't explain why $\mathbb{P}(X_n \geq X + \epsilon)$ and $\mathbb{P}(X_n \leq X - \epsilon) $ are necessarily 0 or 1, and most importantly the "[...] which in turn implies that $|a_n−a|<\epsilon$" needs justification.

Here's how I would do this : first there is no need to split $\{|X_n - X|\ge\epsilon\}$ into two disjoint events as you do.
Simply note that if $X$ and $X_n$ are both degenerate then so is $|X-X_n|$ (why ?) and for any degenerate variable $Z$ and Borel set $B$, $\{w\in\Omega:Z(\omega)\in B\}\in\{\emptyset,\Omega\}$ (why ?).
By hypothesis we have $\mathbb{P}(|X_n-X| \geq \epsilon) \to 0$ and in particular $\mathbb{P}(|X_n-X| \geq \epsilon)<1$ for all large enough $n$, which implies, since $|X-X_n|$ is degenerate, that $\{\omega\in\Omega:|X_n-X|(\omega) \geq \epsilon\}=\emptyset$ for all large enough $n$ (why ?).

Finally we can go back to the definition of $X_n$ and $X$ : on the one hand we know from the above that, for large enough $n$, $|X_n-X|(\omega)<\epsilon $ for all $\omega$, and on the other hand, by definition $X_n(\omega) = a_n$ and $X(\omega) = a$ for all $\omega$, so by combining these two facts we get that for large enough $n$, $|a_n -a|<\epsilon$, as desired.

[The (why ?)'s are little gaps for you to fill]

  • $\begingroup$ Thanks for the detailed answer. I have added an edit to my original post to include the (why) answers. I'd appreciate it if you can comment on those as well. $\endgroup$ Commented Oct 20, 2023 at 8:51

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