In probability theory, a probability density function (pdf), or density of a continuous random variable, is a function that describes the relative likelihood for this random variable to take on a given value. "probability distribution function" may be used when the probability distribution is defined as a function over general sets of values, or it may refer to the cumulative distribution function, or it may be a probability mass function rather than the density. A probability density function is most commonly associated with absolutely continuous univariate distributions. |

$f _{X}$(x) = $\int_{-\infty}^{x} f(t)$ dt

Probability density function has the following properties:

**1)**$f _{X}(x)$ > 0, for all x $\in$ ($-\infty, +\infty$)

**2)**$\int_{-\infty}^{+\infty} f_{X}(x)$ = 1

If a$_{0}, a_{1}, a_{2}$, ....... is a sequence of real numbers and if

A(s) = a$_{0} + a_{1} + a_{2}s^{2}$ + ........

= $\sum_{i = 1}^{\infty}a_{i}s^{i}$

converge in some interval -s$_{0} < s < s_{0}$, when the sequence is infinite then the function A(s) is known as the generating function of the sequence a$_{i}$The variable s has no significance of its own and is introduced to identify a$_{i}$ as the coefficient of s$^{i}$ in the expansion of A(s).

the sequence a$_{i}$ is bounded, then the comparison with the geometric series shows that A(s) converges atleast for |s| < 1. Some of the important steps on probability density function are given below:

**Write down the given values from the question.**

__Step 1:__**Substitute these values in the given equation and find out the PDF.**

__Step 2:__PDF=$\frac{1}{\sigma \sqrt{2\pi }}e^{-\frac{(x-m)^{2}}{2\sigma ^{2}}}$ Given below are some examples on probability density function.

**Example 1:**Two independent random variables X and Y are both normally distributed with means 1 and 2 and standard deviations 3 and 4 respectively. If Z = X - Y, write the probability density function of Z. Also state the median, s.d. and mean of the distribution of Z. Find probability (Z + 1 $\leq$ 0).

**Solution:**Since X $\sim$ N(1,9)and Y $\sim$ N(2,16) are independent,

Z = X - Y $\sim$ N(1 - 2, 9 + 16), i.e., Z = X - Y $\sim$ N(-1, 25)

Hence probability density function of Z is given by

p(z) = $\frac{1}{5\sqrt{2\pi}}exp[-\frac{1}{2}(\frac{z+1}{5})^{2}]$, -$\infty < z < \infty$

For the distribution of Z, Median = Mean = - 1 and s.d. = $\sqrt{25}$ = 5

and P(Z + 1 < 0) = P(Z $\leq$ -1)

= P(U $\leq$ 0) [ U = $\frac{Z + 1}{5}$ $\sim$ N(0,1)]

= 0.5

**Example 2:**a) If log$_{10}$ X is normally distributed with mean 4 and variance 4, find the probability of 1.202 < X < 83180000. (Given log$_{10}$ 1202 = 3.08, log$_{10}$ 8318 = 3.92)

b) log$_{10}$ X is normally distributed with mean 7 ad variance 3, log$_{10}$ Y is normally distributed with mean 3 and variance unity. If the distributions of X and Y are independent, find the probability of 1.202 < ($\frac{X}{Y}$) < 83180000

**Solution:**a) Since log X is a non decreasing function of X, we have

P(1.202 < X < 83180000) =

P(log$_{10}1.202 < log_{10}$ X < 7.92)

= P(0.08 < Y < 7.92)

where Y = log$_{10}$ X $\sim$ N(4,4) (given)

When Y = 0.08, Z = $\frac{0.08 -4 }{2}$ = - 1.96

and when Y = 7.92, Z = $\frac{7.92 - 4}{2}$ = 1.96

Required probability = P(0.08 < Y < 7.92)

= P(-1.96 < Z < 1.96)

= 2P(0 < Z < 1.96) (By symmetry)

= 2 * 0.4750

= 0.9500

b) P[1.202 < $\frac{X}{Y}$ < 83180000]

= P(log$_{10}$ 1.202< log$_{10}$($\frac{X}{Y}$)<log$_{10}$83180000) = (0.08 < U < 7.92),

Where U = log$_{10}$($\frac{X}{Y}$) = log$_{10}$ X - log$_{10}$Y

Since log$_{10}$ X $\sim$ N(7,3)and log$_{10}$Y $\sim$ N(3,1), are independent

log$_{10}$X - log$_{10}$Y $\sim$ N(7 - 3, 3 + 1)

U = (log$_{10}$X - log$_{10}$Y)$\sim$ N(4,4)

Required probability is given by

p = P(0.08 < U < 7.92) where U $\sim$ N(4,4)

= 0.95