FYUG Even Semester Exam, 2025 STATISTICS: Mathematical Statistics (STADSM-252)

Subject: Statistics

Paper Name/Code: Mathematical Statistics / STADSM-252

Semester: 4th Semester (FYUG)

Time: 3 Hours | Full Marks: 70

Pass Marks: 28

UNIT-I

Question 1 (a) [2 Marks]

Define discrete and continuous random variables with illustration.

A discrete random variable is one that can take on only a finite or countably infinite number of distinct values. Illustration: The number of heads obtained in three tosses of a coin (0, 1, 2, or 3).

A continuous random variable is one that can take any value within a specified range or interval. Illustration: The height of students in a class or the lifetime of an electric bulb.

Question 1 (b) [2 Marks]

Define probability mass function (p.m.f.) of a random variable.

For a discrete random variable X, the probability mass function (p.m.f.) is a function P(X=x) that assigns a probability to each possible value x such that P(X=x) >= 0 and the sum of all P(X=x) equals 1.

Question 1 (c) [2 Marks]

State the properties of cumulative distribution function.

F(x) is a non-decreasing function.

F(-infinity) = 0 and F(+infinity) = 1.

F(x) is right-continuous.

Question 2 (a)(i) [4 Marks]

Define p.d.f. and cumulative distribution function of a random variable.

Probability Density Function (p.d.f.): For a continuous random variable X, the p.d.f. f(x) represents the density of probability at point x such that the integral over its entire range is 1.

Cumulative Distribution Function (c.d.f.): Defined as F(x) = P(X <= x). It gives the probability that the random variable X takes a value less than or equal to x.

Question 2 (a)(ii) [6 Marks]

A discrete random variable X has the given probability distribution. Find k, P(X>=6), P(X<6), and P(2

The sum of probabilities must equal 1: 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1.

Equation: 10k^2 + 9k - 1 = 0. Solving gives k = 0.1 (since k cannot be negative).

Value of k: 0.1

P(X>=6): P(X=6) + P(X=7) = (2k^2) + (7k^2 + k) = 9(0.01) + 0.1 = 0.19.

P(X<6): 1 - P(X>=6) = 1 - 0.19 = 0.81.

P(2 Since there are no integers between 2 and 3 for this discrete variable, the probability is 0.

UNIT-II

Question 3 (a) [2 Marks]

Define mathematical expectation.

The mathematical expectation E(X) of a random variable X is the weighted average of all possible values of X, where the weights are the respective probabilities (for discrete) or the p.d.f. (for continuous).

Question 3 (b) [2 Marks]

Prove that E(X^2) >= [E(X)]^2.

Since the variance V(X) = E(X^2) - [E(X)]^2 and variance is always non-negative (V(X) >= 0), it follows directly that E(X^2) >= [E(X)]^2.

Question 4 (a)(i) [5 Marks]

State and prove the additive property of mathematical expectation.

Statement: For any two random variables X and Y, E(X + Y) = E(X) + E(Y), provided the expectations exist.

Proof (Continuous case):
E(X + Y) = Integral Integral (x + y) f(x,y) dx dy
= Integral Integral x f(x,y) dx dy + Integral Integral y f(x,y) dx dy
= Integral x [Integral f(x,y) dy] dx + Integral y [Integral f(x,y) dx] dy
= Integral x f(x) dx + Integral y f(y) dy = E(X) + E(Y).

UNIT-III

Question 5 (a) [2 Marks]

Define moment-generating function (m.g.f.).

The moment-generating function M_x(t) of a random variable X is defined as E(e^(tX)), for all values of t for which the expectation exists.

Question 6 (a)(i) [6 Marks]

Prove that m.g.f. of the sum of independent random variables is equal to the product of their individual m.g.f.s.

Let Z = X + Y, where X and Y are independent.
M_z(t) = E(e^(t(X+Y)))
= E(e^(tX) * e^(tY))
Since X and Y are independent, E(g(X)h(Y)) = E(g(X))E(h(Y)).
M_z(t) = E(e^(tX)) * E(e^(tY)) = M_x(t) * M_y(t).

UNIT-IV

Question 7 (a) [2 Marks]

Define bivariate random variable.

A bivariate random variable (X, Y) is a pair of random variables defined on the same sample space, representing two different characteristics of the same outcome.

Question 8 (b)(ii) [8 Marks]

Given joint p.d.f. f(x,y) = 8xy for 0 < x < y < 1. Find marginal densities and check independence.

Marginal of X: Integral from x to 1 of (8xy) dy = 4x(1 - x^2).

Marginal of Y: Integral from 0 to y of (8xy) dx = 4y^3.

Independence: X and Y are not independent because f(x,y) != f(x)f(y) and the range of x depends on y.

UNIT-V

Question 9 (a) [4 Marks]

State the properties of binomial distribution.

Discrete distribution with parameters n and p.

Mean = np.

Variance = npq.

Mode is the integer value between (n+1)p - 1 and (n+1)p.

Question 10 (b)(i) [5 Marks]

Define Poisson distribution and find its mean.

A discrete random variable X follows Poisson distribution if P(X=x) = [e^(-lambda) * lambda^x] / x! for x = 0, 1, 2...

Mean: E(X) = Sum x * [e^(-lambda) * lambda^x] / x! = lambda.

Exam Focus Enhancements
[span_54](start_span)
Exam Tips: When proving properties of m.g.f. or expectation, always specify if you are using the discrete or continuous case definition.[span_54](end_span)

Common Mistakes: Forgetting that independence of variables requires both the density function to factorize AND the range of one variable to be independent of the other.

Important Formulas List:

E(X) = Sum x P(x)

V(X) = E(X^2) - [E(X)]^2

Binomial m.g.f. (about origin): (q + pe^t)^n

Answer Presentation Strategy: For probability table questions (like 2(a)(ii)), always show the step where you sum probabilities to 1 to justify your value of 'k'.