I can find two definitions of Probability Density Functions in the sources I have checked:

$$P(a < X < b) = \int^b_a f(x)dx$$

Ref: Hogg & Tanis, Probability and Statistical Inference and this from Utah's math dept


$$P(a \leq X \leq b) = \int^b_a f(x)dx$$

Refs: Clarke & Cooke, A Basic Course in Statistics. Wolfram Mathworld: Probability density function

Which is correct? But if one defines the Cumulative Density Function as:

$$P( X \leq a) = D(a) = \int^a_{-\infty} f(x) dx$$


$$D(b) - D(a) = \int^b_a f(x) dx$$


$$P( X \leq b) - P( X \leq a) = \int^b_a f(x) dx = P( a < X \leq b) $$

Or am I making a silly mistake?

  • $\begingroup$ It was mostly @Woodface. I just did the last little bit. $\endgroup$ Commented Mar 8, 2015 at 14:41

2 Answers 2


Both definitions are equal, because if X is a continous variable, the probability for X to take a single value is zero. This is true because an integral with equal upper and lower limits is always equal to zero:

$P(X=a)= \int_a^a f(x)dx= F(a)-F(a) = 0$.

Since the events $\{X=a\}$ and $\{X >a\}$ are distinct, it follows by sigma additivity that:

$P(a \leq X <b)= P(X=a) + P(a<X<b)= P(a<X<b)$.


Statchrist answer is the standard answer for random variables with a probability density function.

However you’re right: if the cdf is $F(x) = P(X\le x)$, then $$P(a < x \le b) = F(b) - F(a),$$ and this is true for all random variables.

If the cdf $F(x)$ is has discontinuities, then $X$ does not have a density, but you can still define integrals $$\int_a^b \phi(x) dF(x)$$ as the limit of the sums $$\sum_{i=0}^{n-1} \phi(c_i)(F(x_{i+1})-F(x_i))$$ with $a = x_0 < \dots < x_n = b$ and $x_i \le c_i \le x_{i+1}$, the limit being taken when $\text{max}(x_{i+1}-x_i) \rightarrow 0$. This is the Stieltjes integral.

With this definition, for all random variables with pdf $F(x)$, $$\int_{a}^b 1\,dF(x) = F(b) - F(a),$$ as all sums of the above form are $\sum_i F(x_{i+1})-F(x_i) = F(b) - F(a)$. Thus, $$\int_{a}^b 1\,dF(x) = F(b) - F(a) = P(a < x \le b),$$ and this quantity can be different to $P(a \le X \le b)$ or $P(a< X<b)$.

  • $\begingroup$ Thank you @Elvis for careful detail, I think I only need to worry about 'well-behaved' situations. $\endgroup$
    – clumb
    Commented Mar 8, 2015 at 10:22
  • 1
    $\begingroup$ You’re right, this was a side note ; this point of view is not standard. However note that this includes the very "well-behaved" situation where $X$ is a discrete variable... this allows to make general statements and proofs without distinguishing between continuous and discrete. In fact I am surprised that Stieltjes integral is not widely used to teach probabilities. $\endgroup$
    – Elvis
    Commented Mar 8, 2015 at 16:19

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