In general the equation that is used in describing a probability distribution that is continuous is termed as a probability density function. In short we can even write PDF or simply a density function.

PDF can define the density function in probability as the
derivative of the distribution function in a continuous distribution.

Let D (x) be the distribution function. Then P (x), which is the density function in probability is given by:

D’ (x) = $[P (x)]^x_{- \infty}$

$\rightarrow$ D’ (x) = P (x) - P (- $\infty$)

$\rightarrow$ D’ (x) = P (x)

Hence when we have X $\leq$ x, then we write

D’ (x) = P (X $\leq$ x) = $\int^x_{- \infty}$ P (e) de

Any
probability function always satisfies P (y $\in$ N) = $\int_N P
(y) dy$ which is contained under the given normalization condition

P (- $\infty$ < m < $\infty$) = $\int_{- \infty}^{\infty}$ P (m) dm = 1.

A density function in probability for any given distribution in probability that is continuous has the following properties:

r (y) = $\left\{\begin{matrix}

y &when\ 0 < y < 1 \\

2-y& when\ 1 < y < 2\\

0& elsewhere \end{matrix}\right.$

y &when\ 0 < y < 1 \\

2-y& when\ 1 < y < 2\\

0& elsewhere \end{matrix}\right.$

It can be verified that int r (y) dy = 1 which shows that ‘r’ is a valid density function. Also we can see that the density of this given function is larger near around 1. If we take up any random variable say ‘Y’, then we can verify easily that

P ($\frac{1}{2}$ < Y < $\frac{3}{2}$) = $\int^{\frac{3}{2}}_{\frac{1}{2}}$ r (y) dy

= $\frac{3}{4}$

The density does not matter at a single point. Hence if we define r (1) = 0 it would not at all matter. Thus we can redefine the given function as below:

y&when\ 0 < y \leq 1 \\

2-y& when\ 1 < y \leq 2\\

0& elsewhere \end{matrix}\right.$

This new defined function is such that we have r (1) = 1. This makes our given function continuous which was continuous. Also the value of the integrand does not depend upon its value at any one point so the value of the random variable ‘Y’ is unchanged even if we change ‘r’ at a point.

**Example 2:** Let Y be a continuous random variable whose PDF is as g(y) = $\frac{y^4}{125}$ for y $\in$ (0, a). Constant "a" makes g(y) a valid PDF. Find the value of a.

**Solution:** Given function is g(y) = $\frac{y^4}{125}$ for y $\in$ (0, a)

To find "a", solve the integral of g(y) having limits 0 and a.

That is $\int_0^a$ $\frac{y^4}{125}$dy = $\frac{1}{125}$ ($\frac{y^5}{5}$)|$_0^a$

= $\frac{1}{625}$ ($a^5$ - 0)

= $\frac{1}{625}$ ($a^5$)

Equate above integral to one.

$\frac{1}{625}$ ($a^5$) = 1

$a^5$ = 625

To find "a", solve the integral of g(y) having limits 0 and a.

That is $\int_0^a$ $\frac{y^4}{125}$dy = $\frac{1}{125}$ ($\frac{y^5}{5}$)|$_0^a$

= $\frac{1}{625}$ ($a^5$ - 0)

= $\frac{1}{625}$ ($a^5$)

Equate above integral to one.

$\frac{1}{625}$ ($a^5$) = 1

$a^5$ = 625