Im a little confused about the difference of these two concepts, especially the convergence of probability. I understand that $X_{n} \overset{p}{\to} Z $ if $Pr(|X_{n} - Z|>\epsilon)=0$ for any $\epsilon >0$ when $n \rightarrow \infty$.

I just need some clarification on what the subscript $n$ means and what $Z$ means. Is $n$ the sample size? is $Z$ a specific value, or another random variable? If it is another random variable, then wouldn't that mean that convergence in probability implies convergence in distribution? Also, Could you please give me some examples of things that are convergent in distribution but not in probability?

  • $\begingroup$ See: quora.com/… $\endgroup$
    – user17900
    Commented Mar 16, 2019 at 15:45
  • $\begingroup$ A quick example: $X_n = (-1)^n Z$, where $Z \sim N(0,1)$. Then $X_n$ does not converge in probability but $X_n$ converges in distribution to $N(0,1)$ because the distribution of $X_n$ is $N(0,1)$ for all $n$. $\endgroup$
    – chan1142
    Commented Apr 28, 2019 at 3:11
  • $\begingroup$ And, no, $n$ is not the sample size. It is just the index of a sequence $X_1,X_2,\ldots$. And $Z$ is a random variable, whatever it may be. In econometrics, your $Z$ is usually nonrandom, but it doesn’t have to be in general. $\endgroup$
    – chan1142
    Commented Apr 28, 2019 at 3:17

1 Answer 1


I will attempt to explain the distinction using the simplest example: the sample mean. Suppose we have an iid sample of random variables $\{X_i\}_{i=1}^n$. Then define the sample mean as $\bar{X}_n$. As the sample size grows, our value of the sample mean changes, hence the subscript $n$ to emphasize that our sample mean depends on the sample size.

Noting that $\bar{X}_n$ itself is a random variable, we can define a sequence of random variables, where elements of the sequence are indexed by different samples (sample size is growing), i.e. $\{\bar{X}_n\}_{n=1}^{\infty}$. The weak law of large numbers (WLLN) tells us that so long as $E(X_1^2)<\infty$, that $$plim\bar{X}_n = \mu,$$ or equivalently $$\bar{X}_n \rightarrow_P \mu,$$

where $\mu=E(X_1)$. Formally, convergence in probability is defined as $$\forall \epsilon>0, \lim_{n \rightarrow \infty} P(|\bar{X}_n - \mu| <\epsilon)=1. $$ In other words, the probability of our estimate being within $\epsilon$ from the true value tends to 1 as $n \rightarrow \infty$. Convergence in probability gives us confidence our estimators perform well with large samples.

Convergence in distribution tell us something very different and is primarily used for hypothesis testing. Under the same distributional assumptions described above, CLT gives us that $$\sqrt{n}(\bar{X}_n-\mu) \rightarrow_D N(0,E(X_1^2)).$$ Convergence in distribution means that the cdf of the left-hand size converges at all continuity points to the cdf of the right-hand side, i.e. $$\lim_{n \rightarrow \infty} F_n(x) = F(x),$$ where $F_n(x)$ is the cdf of $\sqrt{n}(\bar{X}_n-\mu)$ and $F(x)$ is the cdf for a $N(0,E(X_1^2))$ distribution. Knowing the limiting distribution allows us to test hypotheses about the sample mean (or whatever estimate we are generating).

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    $\begingroup$ Your definition of convergence in probability is more demanding than the standard definition. For example, suppose $X_n = 1$ with probability $1/n$, with $X_n = 0$ otherwise. It’s clear that $X_n$ must converge in probability to $0$. However, $X_n$ does not converge to $0$ according to your definition, because we always have that $P(|X_n| < \varepsilon ) \neq 1$ for $\varepsilon < 1$ and any $n$. $\endgroup$ Commented Mar 17, 2019 at 16:59
  • $\begingroup$ Yes, you are right. I posted my answer too quickly and made an error in writing the definition of weak convergence. I have corrected my post. $\endgroup$
    – dlnB
    Commented Mar 17, 2019 at 18:09

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