# What is the relationship betweeen a pdf and cdf?

I am learning stats. On page 20, my book, All of Statistics 1e, defines a CDF as function that maps x to the probability that a random variable, X, is less than x.

$F_{x}(x) = P(X\leq x)$

On page 23 it gives a function

$P(a < X < b ) = \int_{a}^{b}f_{X}dx$

and then says that "the function $f_{X}$ is called the probability density function. We have that..."

$F_{x}(x) = \int_{-\infty}^{x}f_{X}dt$

I am a little confused about how to characterize the most important difference between them. The equation above says that the cdf is the integral of the pdf from negative infinity to x. Is it fair to say that the cdf is the integral of the pdf from negative infinity to x?

• Read carefully, $f_X$ is pdf, $F_x$ is CDF. – booksee May 28 '15 at 3:43
• "The equation above says that the cdf is the integral of the pdf from negative infinity to x. Is it fair to say that the cdf is the integral of the pdf from negative infinity to x?" ??? Indeed it is correct to say that the cdf is the integral of the pdf from negative infinity to x. – Did May 28 '15 at 12:46

Yes. That's correct. A PDF is a probability density function. It is stating the probability of a particular value coming out. Taking this analogy to a discrete distribution, the PDF of a 6-sided die is: $[x<1:0,x=1:\frac{1}{6},x=2:\frac{1}{6},x=3:\frac{1}{6},x=4:\frac{1}{6},x=5:\frac{1}{6},x=6:\frac{1}{6},x>6:0]$. For a continuous probability distribution, you can't really use the PDF directly, since the probability of an infinitesimally thin slice of the PDF being selected is intuitively zero.

That's where the cumulative density function, or CDF, comes it. It is a measure of how likely the value is to be less than some arbitrary value (which we pick). For a discrete case, you start with the first possible value, and add all the entries in the PDF up to the value of interest: $$CDF=\sum PDF \rightarrow [x<1:0,x<2:\frac{1}{6},x<3:\frac{2}{6},x<4:\frac{3}{6},x<5:\frac{4}{6},x<6:\frac{5}{6},x\geq 6:\frac{6}{6}]$$ Notice how the final value of the CDF is $1$. This is expected, since every possible outcome of rolling a 6-sided die is less than or equal to 6.

Now let's go back to the continuous probability distribution. In this case, we don't have a finite set of options for the answer to be, so we can't constrain $X$. Thus, we start from $-\infty$, since that encompasses everything to the left of the chosen $x$. As you should be aware from calculus, the integral is to continuous functions what a sum is to discrete functions - loosely. The value of a CDF is that you can use it to determine the probability of the number falling within a specific range as follows:

$$F(a\leq X \leq b) = F(X \leq b) - F(X \leq a) = \int_{-\infty}^{b} f(x)dx - \int_{-\infty}^{a} f(x)dx = \int_{a}^{b} f(x)dx$$

• The notation for the PDF of the die roll may be a bit confusing. The value of the CDF at $x$ actually is $\frac16$ if $1\leq x < 2$ and $0$ if $x < 1$ (because, for example, $P(X\leq0.99)=0$, but $P(X\leq1.7)=\frac16$). A graph might make the function's values a little clearer. – David K May 28 '15 at 12:55
• Good points. Let me add the entries for which it is zero. – FundThmCalculus May 28 '15 at 12:56
• @FundThmCalculus thanks! Great answer. I'm learning on my own from a book so this kind of explanation is extremely helpful. – bernie2436 May 28 '15 at 22:44
• @bernie2436, thanks for letting me know that. By accepting and upvoting answers, you help the community know what you are looking for. Also, it was a very good question. :) – FundThmCalculus May 29 '15 at 12:01

The cumulative distribution function $F_X$ of any random variable $X$ is defined by $$F_X(x)=P(X\le x)$$ for $x\in\mathbb R$. If $X$ is a continuous random variable, then the cumulative distribution function $F_X$ can be expressed as $$F_X(x)=\int_{-\infty}^xf(x)\mathrm dx$$ for $x\in\mathbb R$, where $f$ is the probability density function of a random variable $X$. If a random variable $X$ is not continuous, it does not have a probability density function (for example, a random variable with Bernoulli distribution).

Simply put, yes, the cdf (evaluated at $x$) is the integral of the pdf from $-\infty$ to $x$. Another way to put it is that the pdf $f(x)$ is the derivative of the cdf $F(x)$.

These definitions assume that the cdf is differentiable everywhere. I mention this not to make the definitions more complicated, but to reduce the factor of surprise later when you find out that there are other kinds of probability distributions that do not have the kind of pdf described here.