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Suppose $\bar{X}_n$ is the mean of a random sample of size ${n}$ from an exponential distribution with $\lambda$ > 0. Then what does the following statement about convergence mean (how does this converge)?

$$\text{exp} \left(-\frac{1}{\bar{X}_n} \right) \xrightarrow{\rm{P}} \text{exp}(-\lambda) $$

More specifically what does the $\rm{P}$ on top of the arrow mean? I understand it's probability but what is the difference between some expression converging to some value in probability versus in distribution?

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As for your last question "what is the difference between some expression converges to some value in terms of distribution and probability?", if by "value" you mean "constant" (for example, the constant ${\rm e}^{-\lambda}$), then there is no difference at all, since convergence in distribution to a constant implies convergence in probability. See…. – Shai Covo Feb 9 '11 at 17:22
I don't think I'm at the level of understanding the proof though I will try but I get the idea. Is there an intuition you can get me for this convergence (when it is the same and when it is not the same) if possible so that I can get a picture of it ? – Sunil Feb 9 '11 at 17:28
@shai: I think you forgot to answer yest but please do whenever you get time. Thanks – Sunil Feb 10 '11 at 15:18
up vote 4 down vote accepted

We say that $X_n\to X$ in distribution if $F_n(x)\to F(x)$ for all continuity points $x$ of $F$. Here, $F_n$ is the distribution function of $X_n$ and $F$ the distribution function of $X$.

On the other hand, $X_n\to X$ in probability means that for every $\varepsilon >0$, we have $$P(|X_n-X|>\varepsilon)\to 0\mbox{ as } n\to\infty.$$

The two concepts are similar, but not quite the same. In fact, convergence in probability is stronger, in the sense that if $X_n\to X$ in probability, then $X_n\to X$ in distribution. It doesn't work the other way around though; convergence in distribution does not guarantee convergence in probability.

As to your question "How?", the answer is that for any continuous function $g$, if $X_n\to X$ then $g(X_n)\to g(X)$. This holds for both convergence in distribution and for convergence in probability.

Since $\bar X_n\to1/\lambda$ in probability by the weak law of large numbers, it follows that $\exp(-1/\bar X_n)\to \exp(-\lambda)$ in probability as $n\to \infty$.

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So for instance $\frac{1}{\bar{X}_n} \xrightarrow{D} \frac{1}{\frac{1}{\lambda}} = \lambda$ ~ N($\lambda, \sigma^2$). Am I right ? I made this question up btw. – Sunil Feb 9 '11 at 16:38
Yes, $1/\bar X_n$ converges in distribution to $\lambda$. But $\lambda$ is a constant and doesn't have a normal distribution unless $\sigma^2=0$. To get central limit behaviour, you need a different scaling: consider instead $\sqrt{n}(\bar X_n-1/\lambda)$. – Byron Schmuland Feb 9 '11 at 16:47
Can you explain what this scaling factor mean and how this helps for $\sigma^2 \rightarrow 0$ ? – Sunil Feb 9 '11 at 17:05

In response to the OP's request, I give two elaborated examples in order to clarify the difference between convergence in probability (denoted by $\stackrel{{\rm P}}{\to}$) and convergence in distribution (denoted by $\stackrel{{\rm D}}{\to}$).

Example 1. Suppose that $X_i$, $i=1,2,\ldots$, are non-constant random variables taking values in $[0,M]$, and let $(a_n)$ be a sequence of positive numbers such that $\sum\nolimits_{i = 1}^\infty {a_i } = c < \infty $. Define $S_n = \sum\nolimits_{i = 1}^n {a_i X_i }$. Then, since the sequence $S_n$ is monotone increasing, $S_n$ converges pointwise (that is, for all $\omega \in \Omega$) to a random variable $S$ taking values in $[0,Mc]$. Here, we have the strongest type of convergence (sure convergence), which implies all the other kinds of convergence. In particular, as one would expect, $S_n \stackrel{{\rm P}}{\to} S$. Indeed, this can be shown directly as follows. Fix $\varepsilon > 0$. Then, for all sufficiently large $n$, $$ {\rm P}(|S_n - S| > \varepsilon) = {\rm P}\bigg(\sum\limits_{i = n + 1}^\infty {a_i X_i } > \varepsilon \bigg) \leq {\rm P}\bigg(M \sum\limits_{i = n + 1}^\infty {a_i } > \varepsilon \bigg) = 0. $$ Now, since convergence in probability implies convergence in distribution, $S_n \stackrel{{\rm D}}{\to} S$ as well. However, the limit random variable $S$ plays no special role with regard to the convergence in distribution. Indeed, take, for example, an independent copy $S'$ of $S$. Then, trivially, $S_n \stackrel{{\rm D}}{\to} S'$ (simply because $S$ and $S'$ have the same distribution function). On the other hand, the limit $S$ plays an essential role with regard to the convergence in probability. In fact, it is easy to prove the following general statement: the limit of convergence in probability is unique in the sense that if $Z_n \stackrel{{\rm P}}{\to} X$ and $Z_n \stackrel{{\rm P}}{\to} Y$, then $X = Y$ almost surely, that is ${\rm P}(X \neq Y) = 0$. Finally, it is worth noting that if $Z_n \stackrel{{\rm D}}{\to} Z$, and $Z$ is distributed according to a distribution $F$, then we can write $Z_n \stackrel{{\rm D}}{\to} F$. For example, if $Z_n \stackrel{{\rm D}}{\to} Z$ where $Z \sim {\rm exponential}(\lambda)$, then we can write $Z_n \stackrel{{\rm D}}{\to} {\rm exponential}(\lambda)$.

To further clarify the difference between convergence in probability and convergence in distribution, let's consider the fundamental case of the central limit theorem.

Example 2. Suppose that $X_1,X_2,\ldots$ is a sequence of i.i.d. random variables with expectation $\mu$ and (finite) variance $\sigma^2 > 0$. Define $S_n = X_1 + \cdots + X_n$ and $Z_n = \frac{{S_n - n\mu }}{{\sigma \sqrt n }}$. The central limit theorem states that $Z_n$ converges in distribution to the standard normal distribution ${\rm N}(0,1)$, that is $Z_n \stackrel{{\rm D}}{\to} {\rm N}(0,1)$. So, given any random variable $Z \sim {\rm N}(0,1)$ (which, in particular, may be defined on a different probability space), we can write $Z_n \stackrel{{\rm D}}{\to} Z$. On the other hand, there is no random variable $Z$ such that $Z_n \stackrel{{\rm P}}{\to} Z$. Indeed, suppose for a contradiction that $Z_n \stackrel{{\rm P}}{\to} Z$. It is an easy exercise to show that $$ {\rm P}(|Z_n - Z_m | > 2 \varepsilon ) \le {\rm P}(|Z_n - Z| > \varepsilon ) + {\rm P}(|Z_m - Z| > \varepsilon ). $$ Now, in order to reach a contradiction, it suffices to realize that $Z_n$ and $Z_m$ become asymptotically independent as $n,m \to \infty$ with $n/m \to 0$; indeed, $$ Z_m = \sqrt {\frac{n}{m}} Z_n + \sqrt {\frac{{m - n}}{m}} \frac{{\sum\nolimits_{i = n + 1}^m {X_i } - (m - n)\mu }}{{\sigma \sqrt {m - n} }}, $$ from which it is also seen that $$ {\rm Cov}(Z_n,Z_m) = \sqrt {\frac{n}{m}}. $$

Finally, especially in view of the first example, it is worth noting that convergence in probability, though quite strong relative to convergence in distribution, does not imply almost sure convergence. A short but sophisticated example is given in my answer to this question, at the end of the second paragraph.

EDIT: As I commented below, I intentionally gave the non-trivial example of the central limit theorem. Here are two trivial examples.

First (elaborating on Didier's example), if $X_1,X_2,\ldots$ are i.i.d. from a distribution $F$, then, trivially, $X_n \stackrel{{\rm D}}{\to} F$ (since $X_i \sim F$ for each $i$). But, unless the $X_i$ are deterministic, the sequence never converges in probability. Indeed, suppose that $X_n \stackrel{{\rm P}}{\to} X$. Let $\varepsilon >0 $ be arbitrary but fixed. By the triangle inequality, the event $\lbrace |X_{n+1} - X_n| > 2 \varepsilon \rbrace$ is contained in the event $\lbrace |X_{n+1} - X| > \varepsilon \rbrace \cup \lbrace |X_{n} - X| > \varepsilon \rbrace$. Hence, $$ {\rm P}(|X_{n+1}-X_n| > 2 \varepsilon) \leq {\rm P}(|X_{n + 1} - X| > \varepsilon ) + {\rm P}(|X_n - X| > \varepsilon ). $$ Since, by our assumption, the right-hand side tends to $0$ as $n \to \infty$, and since $|X_{n+1}-X_n|$ is equal in distribution to $Y:=|X_1 - X_2|$, we get ${\rm P}(Y > 2 \varepsilon) = 0$. Since $Y$ is nonnegative, this implies that $Y = 0$ almost surely (exercise), that is, $X_1 = X_2$ almost surely. Hence, the $X_i$ are deterministic (since they are independent).

As another example, suppose that ${\rm P}(X_1 = 1) = {\rm P}(X_1 = -1) = 1/2$, and let $X_{n+1}=-X_n$. Then, trivially, $X_n \stackrel{{\rm D}}{\to} X_1$, but either $(X_n) = (1,-1,1,-1,\ldots)$ or $(X_n) = (-1,1,-1,1,\ldots)$.

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Thank you for your help. I highly appreciate it. Now I get the idea. When explained with examples it's more clear to understand for those who are new to the statistics field. – Sunil Feb 14 '11 at 5:54
And Example 3 (or should it be Example 0?): consider any i.i.d. sequence $(X_i)$. Then $(X_i)$ converges in distribution. In fact, regarding convergence in distribution, $(X_i)$ is like a constant sequence. But, unless the $X_i$ are deterministic in the sense that there exists $x$ such that $P(X_i=x)=1$, $(X_i)$ never converges in probability. You can try to prove this first for Bernoullis if this helps. – Did Feb 14 '11 at 7:06
@Didier: I intentionally gave the non-trivial example of CLT, because of its fundamental importance (see also the comments below Byron's answer). Anyway, I'll add later on a couple of trivial examples, including yours "Example 3". – Shai Covo Feb 14 '11 at 10:29
Sure, no problem. – Did Feb 14 '11 at 11:31

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