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Textbook says that for continuous r.v. the probability at a specific value is zero:

$$P(X=a) = 0$$

They say that the proof of this fact is:

$$P\Big(a\Big) \le P\Big((a-\epsilon) \le x \lt a\Big)$$

(because "a" is a subset of the range $(a-\epsilon)$ to "a").

Then if we allow epsilon to approaches zero:

$$\lim \limits_{\epsilon \to 0} P\Big((a-\epsilon) < x \lt a\Big) = \lim \limits_{\epsilon \to 0} \Big(F_X(a) - F_X(a-\epsilon)\Big)$$

therefore:

$$P(X= a) = 0$$

(note: $F_X(x) = P(X \le a)$ is the cdf of pdf $f_X(x)$)

However, a paragraph later in the textbook, it says that the pdf of continuous r.v. has the property:

$$f_{X}(x) \ge 0$$

My question is this: What's the difference between $P(X=a)$ and $f_X(a)$? shouldn't it be : $f_{X}(a) = 0$

without the greater than sign?

pico
  • 941

2 Answers2

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A continuous RV has no probability mass function (PMF) $p$ since a requirement for a PMF is that the sum of the PMF values is $1$. Attempting to define a PMF $p$ for a continuous RV $X$ leads to $p(a)=\int_a ^a f(x)\text{d}x=0.$ Therefore, we cannot obtain $\sum_{-\infty < a<\infty}p(a)=1$ for a continuous $X$.

Also, we cannot have $f(x)=0$ for all $x$ since (letting $S$ denote the sample space) we need $$\mathbb{P}(X\in S)=\int_{\mathbb R}f(x)\text{d}x=1.$$

The analogue of the PMF for a continuous RV $X$ is a function (PDF) $f$.

Continuous case:

  • $f\ge 0$,
  • $\mathbb P(X \in A)=\int_A f(x)\text{d}x$,
  • $\int_{-\infty}^\infty f(x)\text{d}x=1$

Discrete case:

  • $p\ge 0$,
  • $\mathbb{P}(X\in A)=\sum_{x\in A}p(a),$
  • $\sum_{-\infty<a<\infty}p(a)=1.$
The Substitute
  • 3,888
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For discrete r.v:

$$P(X=a) = f_X(a)$$

However, for continuous r.v.:

$$P(X=a) \ne f_X(a)$$

It equals:

$$P(X=a) = \lim \limits_{\epsilon \to 0} \int \limits^{a+\epsilon}_{a}f_x(v)~dv = 0$$

because:

$$P(a<X<b) = \int \limits^{b}_{a}f_x(v)~ dv = 0$$

pico
  • 941