Probability – - Expectation Of Binomial Distribution

Probability – Expectation of Binomial Distribution

Binomial distribution: a discrete probability distribution
Involves a fixed number of trials, each with 2 possible outcomes: success or failure
Each trial is independent and has the same probability of success

Notation used in Binomial Distribution

n: number of trials
p: probability of success in each trial
q: probability of failure in each trial (q = 1 - p)
X: random variable representing the number of successes in the n trials

Formula for Expectation of Binomial Distribution

The expectation (mean) of a binomial distribution is given by:

E(X) = n * p This represents the average number of successes in n trials.

Example: Expectation of Binomial Distribution

Suppose we have a fair coin that we flip 10 times. Let’s find the expectation.

n = 10 (number of trials)
p = 0.5 (probability of success: getting heads) Using the formula E(X) = n * p:
E(X) = 10 * 0.5 = 5 Therefore, the expectation is 5, which means we can expect to get 5 heads on average when flipping the coin 10 times.

Properties of Expectation of Binomial Distribution

The expectation of a binomial distribution is always a whole number
The expectation may or may not be equal to any of the possible values of X
The expectation is affected by both n and p

Variance of Binomial Distribution

The variance of a binomial distribution measures how the values of X are scattered around the expectation
Variance represents the average of the squared deviations from the mean

Formula for Variance of Binomial Distribution

The variance of a binomial distribution is given by:

Var(X) = n * p * q This represents the average squared deviation from the mean.

Example: Variance of Binomial Distribution

Continuing from the previous example, let’s find the variance of the binomial distribution.

n = 10 (number of trials)
p = 0.5 (probability of success: getting heads)
q = 1 - 0.5 = 0.5 (probability of failure: getting tails) Using the formula Var(X) = n * p * q:
Var(X) = 10 * 0.5 * 0.5 = 2.5 Therefore, the variance is 2.5.

Standard Deviation of Binomial Distribution

The standard deviation of a binomial distribution is the square root of the variance
It measures the spread or dispersion of the values of X

Formula for Standard Deviation of Binomial Distribution

Standard Deviation = √(n * p * q) This represents the average spread or dispersion of the values of X.

Properties of Standard Deviation

The standard deviation of a binomial distribution can never be negative
The standard deviation is affected by both n and p
A larger value of n leads to a larger standard deviation
A smaller value of p or q leads to a larger standard deviation

Example: Standard Deviation of Binomial Distribution

Continuing from the previous example, let’s find the standard deviation of the binomial distribution.

n = 10 (number of trials)
p = 0.5 (probability of success: getting heads)
q = 1 - 0.5 = 0.5 (probability of failure: getting tails) Using the formula Standard Deviation = √(n * p * q):
Standard Deviation = √(10 * 0.5 * 0.5) = √(2.5) ≈ 1.58 Therefore, the standard deviation is approximately 1.58.

Expected Value of a Function of a Random Variable

We can also find the expectation of a function of a random variable
Let g(X) be a function of the random variable X
The expectation of g(X), denoted E(g(X)), is given by the formula: E(g(X)) = ∑(g(x) * P(X = x))
Here, x represents each possible value of X and P(X = x) represents the probability of getting that value of X

Example: Expected Value of a Function of a Random Variable

Suppose we have a biased coin that we flip 3 times. Let’s find the expectation of the function g(X) = X^2.

n = 3 (number of trials) Using the formula E(g(X)) = ∑(g(x) * P(X = x)):
When X = 0: g(X) = 0^2 = 0, P(X = 0) = (1/2)^3 = 1/8
When X = 1: g(X) = 1^2 = 1, P(X = 1) = 3 * (1/2)^3 = 3/8
When X = 2: g(X) = 2^2 = 4, P(X = 2) = 3 * (1/2)^3 = 3/8
When X = 3: g(X) = 3^2 = 9, P(X = 3) = (1/2)^3 = 1/8 E(g(X)) = (0 * 1/8) + (1 * 3/8) + (4 * 3/8) + (9 * 1/8) = 16/8 = 2 Therefore, the expectation of g(X) = X^2 is 2.

Mean of a Continuous Random Variable

In the case of a continuous random variable, the mean is called the expected value or expected mean
Denoted by E(X) or μ
Calculated by integrating the product of the random variable X and its probability density function (PDF) over its entire range E(X) = ∫(x * f(x)) dx
Here, f(x) represents the probability density function of X

Variance of a Continuous Random Variable

Variance measures the spread or dispersion of a continuous random variable
Denoted by Var(X) or σ^2
Calculated by integrating the squared deviation of X from its mean, weighted by the probability density function (PDF), over the entire range of X Var(X) = ∫((x - E(X))^2 * f(x)) dx
Here, f(x) represents the probability density function of X

Uniform Distribution

A continuous random variable follows a uniform distribution if it has a constant PDF over a specified interval
The PDF of a uniform distribution is given by: f(x) = 1 / (b-a), a ≤ x ≤ b
The expected value (mean) of a uniform distribution is given by: E(X) = (a + b) / 2
The variance of a uniform distribution is given by: Var(X) = (b - a)^2 / 12

Example: Uniform Distribution

Suppose we have a uniform distribution over the interval [2, 8]. Let’s find the mean and variance.

a = 2 (lower bound)
b = 8 (upper bound) Using the formulas:
E(X) = (2 + 8) / 2 = 5
Var(X) = (8 - 2)^2 / 12 = 36 / 12 = 3 Therefore, the mean is 5 and the variance is 3.

Normal Distribution

One of the most important probability distributions in statistics
Often referred to as the bell curve or Gaussian distribution
It is characterized by its mean (μ) and standard deviation (σ)
The PDF of a normal distribution is given by: f(x) = (1 / sqrt(2πσ^2)) * e^(-(x - μ)^2 / (2σ^2))
The mean (expected value) and variance of a normal distribution are equal to μ and σ^2, respectively

Example: Normal Distribution

Suppose we have a normal distribution with a mean of 70 and a standard deviation of 5. Let’s find the probability of a random variable falling within certain ranges.

P(65 ≤ X ≤ 75) represents the probability of X being between 65 and 75
Using the Z-score formula: Z = (X - μ) / σ Z1 = (65 - 70) / 5 = -1 Z2 = (75 - 70) / 5 = 1
Using a Z-table or technology, we can find the area under the curve between -1 and 1, which represents the probability Therefore, P(65 ≤ X ≤ 75) ≈ 0.6826 (approximately 68.26%).

Conditional Probability

Conditional probability is the probability of an event A occurring, given that event B has already occurred
Conditional probability is denoted by P(A|B)
The formula for conditional probability is: P(A|B) = P(A ∩ B) / P(B)
P(A ∩ B) represents the probability of both A and B occurring

Example: Conditional Probability

Suppose we have a deck of cards with 52 cards. Let A be the event of drawing a red card, and let B be the event of drawing a diamond.

P(A) = 26/52 = 1/2 (probability of drawing a red card)
P(B) = 13/52 = 1/4 (probability of drawing a diamond)
P(A ∩ B) = 13/52 = 1/4 (probability of drawing a red diamond) Using the formula P(A|B) = P(A ∩ B) / P(B):
P(A|B) = (1/4) / (1/4) = 1 Therefore, given that a diamond was drawn, the probability of drawing a red card is 1.

Law of Total Probability

The law of total probability allows us to find the probability of an event A by considering all the possible outcomes of another event B
This law is based on the concept of partitioning or dividing the sample space into mutually exclusive events P(A) = P(A|B1) * P(B1) + P(A|B2) * P(B2) + … + P(A|Bn) * P(Bn)
B1, B2, …, Bn represent mutually exclusive events that partition the sample space, and P(B1), P(B2), …, P(Bn) represent their respective probabilities

Bayes’ Theorem

Bayes’ theorem allows us to update the probability of an event A based on the occurrence of another event B P(A|B) = (P(B|A) * P(A)) / P(B)
P(A|B) represents the probability of event A given that event B has occurred
P(B|A) represents the probability of event B given that event A has occurred
P(A) and P(B) represent the probabilities of events A and B, respectively

Example: Bayes’ Theorem

Suppose a drug test is known to correctly identify 95% of drug users (true positive rate) and correctly identify 90% of non-users (true negative rate). The rate of drug users in the general population is 2%. Let A be the event of a positive drug test and B be the event of being a drug user.

P(A|B) = 0.95 (probability of positive test given drug user)
P(B) = 0.02 (probability of being a drug user)
P(B|A) = ? (the probability of being a drug user given a positive test) Using Bayes’ theorem: P(B|A) = (P(A|B) * P(B)) / P(A)
P(B|A) = (0.95 * 0.02) / P(A) To find P(A), we need to consider both true positives and false positives.

Relations and Functions

A relation is a set of ordered pairs that establishes a relationship between two sets
A function is a special type of relation in which each element of the first set (domain) corresponds to exactly one element of the second set (codomain)
Functions are denoted by f(x), where x is the input and f(x) is the output

Types of Functions

One-to-One (Injective) Function: Each element of the domain corresponds to a unique element in the codomain, and vice versa
Onto (Surjective) Function: Each element of the codomain has at least one corresponding element in the domain
One-to-One Correspondence (Bijective) Function: A function that is both one-to-one and onto

Composite Functions

A composite function is formed by combining two functions, denoted as (f ∘ g)(x), where f and g are functions and x is the input
The composite function (f ∘ g)(x) means applying g(x) first and then applying f(x)
The domain of the composite function is the set of all inputs for which both g(x) and f(x) are defined

Inverse Functions

An inverse function undoes the effect of a given function
If f(x) and g(x) are inverse functions, then (f ∘ g)(x) = x and (g ∘ f)(x) = x for all values in their respective domains
The inverse function is denoted as f^(-1)(x)

Example: Composite and Inverse Functions

Let f(x) = 2x + 3 and g(x) = x - 2. Find the composite function (f ∘ g)(x) and the inverse function f^(-1)(x).

(f ∘ g)(x) = f(g(x)) = f(x - 2) = 2(x - 2) + 3 = 2x - 1 To find the inverse function f^(-1)(x):
Step 1: Replace f(x) with y: y = 2x + 3
Step 2: Swap x and y: x = 2y + 3
Step 3: Solve for y: y = (x - 3) / 2 Therefore, the inverse function is f^(-1)(x) = (x - 3) / 2.