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There is a question from lecture notes whose answer I do not quite understand. The question is

Let $X$ have a cdf $F$. Let $x$ be any point and suppose $x_n$ is a decreasing sequence such that $x_n \rightarrow x$ as $n \rightarrow \infty$. Show $F(x_n) \rightarrow F(x)$ as $n \rightarrow \infty$.

The solution according to lecture notes is:

If $x_n \downarrow -\infty$ Then $\lim_{m\to\infty}F(x_m) = \lim_{m\to\infty}P(X \leq x_m) = P(\lim_{m\to\infty}(X \leq x_m)) = P(\bigcap_{m}(X \leq x_m)) = P(\varnothing) = 0$ Hence $\lim_{x\to -\infty}F(x) = 0$ which can be written as $F(-\infty) = 0$. Then this implies the conclusion.

Can someone help me understand how this argument answers the question or suggest an alternative answer?

mathnoob
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    Simply replace $x_n \downarrow -\infty$ with $x_n \downarrow x$, and note $\cap_m {X \leq x_m} = {X \leq x}$. Also you should check that the result does not hold for $x_n \uparrow x$. – gogurt Jul 02 '13 at 00:02

2 Answers2

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The way you've phrased it, it could work at most for $x=-\infty$.

Probability (at least the way it is conventionally axiomatized) is countably additive. That means if $A_1,A_2,A_3,\ldots$ are a pairwise disjoint sequence of events, then $P(A_1\cup A_2\cup A_3\cup\cdots)$ $=P(A_1)+P(A_2)+P(A_3)+\cdots$.

Finite additivity tells us that $$ \begin{align} F(a)+P(a<X\le b) & =P(X\le a)+P(a<X\le b) \\[8pt] & = P(X\le a\text{ or }a<X\le b)\quad\text{(finite additivity)} \\[8pt] & = P(X\le b) \\[8pt] & = F(b), \end{align} $$ whence $$ P(a<X\le b)=F(b)-F(a). $$

Now suppose $x_n\downarrow x$. Then $$ \begin{align} F(x) & = P(X\le x) = 1 - P(X>x) \\[8pt] & = 1-P\Big( X>x_1\text{ or }x_1\ge X>x_2\text{ or }x_2>X\ge x_3\text{ or }\cdots \Big) \\[8pt] & = 1 - P(X>x_1) - \sum_{i=1}^\infty P(x_i>X\ge x_{i+1})\quad\text{(countable additivity)} \\[8pt] & = F(x_1) - \sum_{i=1}^\infty (F(x_i)-F(x_{i+1})) \\[8pt] & = F(x_1) - \lim_{n\to\infty} \sum_{i=1}^n (F(x_i)-F(x_{i+1})) \\[8pt] & = \lim_{n\to\infty} F(x_1)-F(x_1)+F(x_2)-F(x_2)+\cdots+F(x_{n-1})-F(x_{n-1})+F(x_n) \\[8pt] & = \lim_{n\to\infty}F(x_n). \end{align} $$

Michael Hardy
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Cumulative distribution functions are necessarily continuous since they are the integral of a probability density function. So, we have that $F$ is a continuous function which satisfies the theorem from general topology that if $x_{n} \rightarrow x$ as $n \rightarrow \infty$ and $F$ is continuous, then $F(x_{n}) \rightarrow F(x)$ as $n \rightarrow \infty$. This is a much more natural and intuitive solution than the one presented in your lecture notes.

Samuel Reid
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  • The question is asking to prove the right-continuous property that you are assuming here... – gogurt Jul 01 '13 at 23:51
  • @gogurt: Well, then accept the proof given in the lecture notes. I'm not sure what the problem is... an "I don't understand" type of question is hard to answer without giving an alternative explanation. – Samuel Reid Jul 01 '13 at 23:54
  • You must have meant "integral of a probability density function". But the problem is, not all of them are. – Michael Hardy Jul 02 '13 at 00:02