Mathematics | Probability Distributions Set 1


Prerequisite – Random Variable

In probability theory and statistics, a probability distribution is a mathematical function that can be thought of as providing the probabilities of occurrence of different possible outcomes in an experiment. For instance, if the random variable X is used to denote the outcome of a coin toss (“the experiment”), then the probability distribution of X would take the value 0.5 for X = heads, and 0.5 for X = tails (assuming the coin is fair).
Probability distributions are divided into two classes –

  1. Discrete Probability Distribution – If the probabilities are defined on a discrete random variable, one which can only take a discrete set of values, then the distribution is said to be a discrete probability distribution. For example, the event of rolling a die can be represented by a discrete random variable with the probability distribution being such that each event has a probability of \:\frac{1}{6}.
  2. Continuous Probability Distribution – If the probabilities are defined on a continuous random variable, one which can take any value between two numbers, then the distribution is said to be a continuous probability distribution. For example, the temperature throughout a given day can be represented by a continuous random variable and the corresponding probability distribution is said to be continuous.

Cumulative Distribution Function –
Similar to the probability density function, the cumulative distribution function F(x) of a real-valued random variable X, or just distribution function of X evaluated at x, is the probability that X will take a value less than or equal to x.
For a discrete Random Variable,
 F(x) = P(X\leq x) = \sum \limits_{x_0\leq x} P(x_0)
For a continuous Random Variable,
 F(x) = P(X\leq x) = \int \limits_{-\infty}^{x} f(x)dx

Uniform Probability Distribution –

The Uniform Distribution, also known as the Rectangular Distribution, is a type of Continuous Probability Distribution.
It has a Continuous Random Variable X restricted to a finite interval [a,b] and it’s probability function f(x) has a constant density over this interval.
The Uniform probability distribution function is defined as-

     \[ f(x) =  \begin{cases} \frac{1}{b-a}, & a\leq x \leq b\\ 0, & \text{otherwise}\\ \end{cases} \]

Uniform Distribution graph

Expected or Mean Value – Using the basic definition of Expectation we get –

     \begin{align*} E(x) &= \int \limits_{-\infty}^{\infty} xf(x) dx&\\ &= \int \limits_{a}^{b} \frac{x}{b-a} dx&\\ &= \frac{1}{b-a} \int \limits_{a}^{b} x dx&\\ &= \frac{1}{b-a} \Big[ \frac{x^2}{2}\Big]_{a}^{b}&\\ &= \frac{b^2 - a^2}{2(b-a)}&\\ &= \frac{b + a}{2}&\\ \end{align*}

Variance- Using the formula for variance- V(X) = E(X^2) - (E(X))^2

     \begin{align*} E(x^2) &= \int \limits_{-\infty}^{\infty} x^2f(x) dx&\\ &= \int \limits_{a}^{b} \frac{x^2}{b-a} dx&\\ &= \frac{1}{b-a} \int \limits_{a}^{b} x^2 dx&\\ &= \frac{1}{b-a} \Big[ \frac{x^3}{3}\Big]_{a}^{b}&\\ &= \frac{b^3 - a^3}{3(b-a)}&\\ &= \frac{b^2 + a^2 + ab}{3}&\\ \end{align*}

Using this result we get –

     \begin{align*} V(x) &= \frac{b^2 + a^2 + ab}{3} - \Big( \frac{b+a}{2}\Big) ^2 &\\ &= \frac{b^2 + a^2 + ab}{3} - \frac{b^2+a^2+2ab}{4} &\\ &= \frac{4b^2 + 4a^2 + 4ab - 3b^2 - 3a^2 - 6ab}{12}&\\ &= \frac{(b-a)^2}{12}&\\ \end{align*}

Standard Deviation – By the basic definition of standard deviation,

     \begin{align*} \sigma &= \sqrt{V(x)} \\&= \frac{b-a}{2\sqrt{3}} \end{align*}

  • Example 1 – The current (in mA) measured in a piece of copper wire is known to follow a uniform distribution over the interval [0, 25]. Find the formula for the probability density function f(x) of the random variable X representing the current. Calculate the mean, variance, and standard deviation of the distribution and find the cumulative distribution function F(x).
  • Solution – The first step is to find the probability density function. For a Uniform distribution, f(x) = \frac{1}{b-a}, where b,\:a are the upper and lower limit respectively.

         \therefore  \[  f(x) =  \begin{cases}  \frac{1}{25-0} = 0.04, & 0\leq x\leq 25 \\  0, & \text{otherwise} \\ \end{cases} \]

    The expected value, variance, and standard deviation are-
     E(x) = \frac{b+a}{2} = \frac{25+0}{2} = 12.5 mA\\\\ V(X) = \frac{(b-a)^2}{12} = \frac{(25-0)^2}{12} = 52.08 mA^2\\\\ \text{Standard Deviation} = \sigma = \sqrt{V(x)} = \frac{25}{2\sqrt{3}} = 7.21 mA
    The cumulative distribution function is given as-
     F(x) = \int \limits_{-\infty}^{x} f(x) dx
    There are three regions where the CDF can be defined, x<0,\: 0\leq x\leq 25,\:25 < x

         \[ F(x) =  \begin{cases} 0, &x<0\\ \frac{x}{25}, &0\leq x\leq 25\\ 1, &25<x \end{cases} \]

References –

Probability Distribution – Wikipedia
Uniform Probability Distribution –

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