Random Variables, Expectation, Variance and Covariance PYQs for GATE CS

Solve 6+ Random Variables, Expectation, Variance and Covariance previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Probability and Statistics

Chapter Journey: Probability and Statistics

1. Random Variables and Distributions

Discrete vs Continuous, PMF vs PDF, CDF. Map outcomes to numbers.

2. Expectation and Moments (Current Topic)

Mean, Variance, Covariance. Calculate expected values for geometric quantities.

3. Standard Distributions

Binomial, Poisson, Normal, Exponential. Recognize patterns and apply formulas.

4. Limit Theorems

Law of Large Numbers, Central Limit Theorem. Behavior of sums.

Exam Insight: Questions often blend probability with algorithms (expected running time) or systems (queueing delays). Mastering expectation is non-negotiable.

Core Idea: Geometric Expectation via Integration

The Core Idea

In many problems, a random experiment defines a geometric object (like a sub-interval). We are asked to find the Expected Value (mean) of a property of that object (like its length).

General Method

If $X$ is a continuous random variable with PDF $f_{X} (x)$ , and $Y = g (X)$ is a geometric quantity derived from $X$ , then:

E [Y] = \int_{- \infty}^{\infty} g (x) f_{X} (x) d x

Key Steps

Identify the Random Variable: Usually the position of a cut or a point, e.g., $X \sim U (0, 1)$ .
Define the Geometry: Express the target quantity (length, area) as a function $g (X)$ . Note that this function might be piecewise.
Determine the PDF: For uniform distributions over $(a, b)$ , $f_{X} (x) = \frac{1}{b - a}$ .
Integrate: Compute $\int g (x) f_{X} (x) d x$ over the valid range.

Random Variables, Expectation, Variance and Covariance: Solved Questions with Step-by-Step Explanations (6 Problems)

Question 1 · Engineering Mathematics · 2026_Set1 NAT

Let

X

be a random variable which takes values in the set

{1, 2, 3, 4, 5, 6, 7, 8}

. Further,

Pr (X = 1) = Pr (X = 2) = Pr (X = 5) = Pr (X = 7) = \frac{1}{6}

and

Pr (X = 3) = Pr (X = 4) = Pr (X = 6) = Pr (X = 8) = \frac{1}{12}

.

The expected value of

X

, denoted by

E [X]

, is equal to ___________. (rounded off to two decimal places)

Correct Answer:

4.25

Step-by-Step Solution

Insight: The expected value of a discrete random variable is the sum of each value multiplied by its probability. Grouping terms with the same probability simplifies the arithmetic.

Exam route: Group the values by probability. Sum of values with $p = 1/6$ is $1 + 2 + 5 + 7 = 15$ . $15 \times (1/6) = 2.5$ . Sum of values with $p = 1/12$ is $3 + 4 + 6 + 8 = 21$ . $21 \times (1/12) = 1.75$ . Total $E [X] = 2.5 + 1.75 = 4.25$ .

Learning route:

Recall the definition of discrete expectation: $E [X] = \sum x_{i} P (X = x_{i})$ .
Group the outcomes by their probabilities to simplify calculation:

Group 1 ( $p = 1/6$ ): $x \in {1, 2, 5, 7}$ . Sum of $x = 1 + 2 + 5 + 7 = 15$ .
Group 2 ( $p = 1/12$ ): $x \in {3, 4, 6, 8}$ . Sum of $x = 3 + 4 + 6 + 8 = 21$ .

Multiply each group's sum by its probability:

Contribution of Group 1: $15 \times \frac{1}{6} = 2.5$ .
Contribution of Group 2: $21 \times \frac{1}{12} = \frac{7}{4} = 1.75$ .

Add the contributions: $E [X] = 2.5 + 1.75 = 4.25$ .

Question 2 · Engineering Mathematics · 2025_Set2 NAT

The unit interval

(0, 1)

is divided at a point chosen uniformly distributed over

(0, 1)

R

into two disjoint subintervals.

The expected length of the subinterval that contains 0.4 is ___________. (rounded off to two decimal places)

Correct Answer:

0.74

Step-by-Step Solution

Insight: The length of the subinterval containing a fixed point $c$ depends on whether the random cut $X$ is to the left or right of $c$ . This requires a piecewise function and splitting the integral.

Exam route: Use the derived formula for the expected length containing $c$ in a unit interval: $E [L] = c - c^{2} + 0.5$ . For $c = 0.4$ , $E [L] = 0.4 - 0.16 + 0.5 = 0.74$ .

Learning route:

Define the random variable: Let $X \sim U (0, 1)$ be the cut point. The PDF is $f_{X} (x) = 1$ for $0 < x < 1$ .
Define the target variable $L$ (length containing $c = 0.4$ ):

If $X < 0.4$ , the interval containing $0.4$ is $(X, 1)$ , so $L = 1 - X$ .
If $X > 0.4$ , the interval containing $0.4$ is $(0, X)$ , so $L = X$ .

Set up the expectation integral:

$E [L] = \int_{0}^{0.4} (1 - x) (1) d x + \int_{0.4}^{1} (x) (1) d x$ .

Evaluate the integrals:

First part: $[x - x^{2} /2]_{0}^{0.4} = 0.4 - 0.08 = 0.32$ .
Second part: $[x^{2} /2]_{0.4}^{1} = 0.5 - 0.08 = 0.42$ .

Sum the results: $0.32 + 0.42 = 0.74$ .

Question 3 · Engineering Mathematics · 2024_Set2 MCQ

Let $x$ and $y$ be random variables, not necessarily independent, that take real values in the interval $[0, 1]$ . Let $z = x y$ and let the mean values of $x$ , $y$ , $z$ be $\overset{x}{ˉ}$ , $\overset{y}{ˉ}$ , $\overset{z}{ˉ}$ , respectively. Which one of the following statements is TRUE?

A.
$\overset{z}{ˉ} = \overset{x}{ˉ} \overset{y}{ˉ}$
B.
$\overset{z}{ˉ} \leq \overset{x}{ˉ} \overset{y}{ˉ}$
C.
$\overset{z}{ˉ} \geq \overset{x}{ˉ} \overset{y}{ˉ}$
D.
$\overset{z}{ˉ} \leq \overset{x}{ˉ}$

Correct Answer:

Step-by-Step Solution

Insight: For random variables bounded in $[0, 1]$ , the product $X Y$ is always less than or equal to $X$ (since $Y \leq 1$ ) and less than or equal to $Y$ (since $X \leq 1$ ).

Exam route: Since $0 \leq Y \leq 1$ , we have $X Y \leq X$ . By monotonicity of expectation, $E [X Y] \leq E [X]$ , which means $\overset{z}{ˉ} \leq \overset{x}{ˉ}$ . This matches option D.

Learning route:

Analyze the bounds: We are given $X, Y \in [0, 1]$ .
Establish inequality for the product: Since $Y \leq 1$ and $X \geq 0$ , multiplying by $X$ gives $X Y \leq X$ .
Apply expectation: The expectation operator preserves inequalities. Thus, $E [X Y] \leq E [X]$ .
Translate to given notation: $\overset{z}{ˉ} \leq \overset{x}{ˉ}$ .
Check other options: $\overset{z}{ˉ} = \overset{x}{ˉ} \overset{y}{ˉ}$ is only true if independent. $\overset{z}{ˉ} \leq \overset{x}{ˉ} \overset{y}{ˉ}$ and $\overset{z}{ˉ} \geq \overset{x}{ˉ} \overset{y}{ˉ}$ are not universally true without knowing the correlation (covariance sign).

Question 4 · Engineering Mathematics · 2021_Set2 MCQ

In an examination, a student can choose the order in which two questions (QuesA and QuesB) must be attempted.
- If the first question is answered wrong, the student gets zero marks.
- If the first question is answered correctly and the second question is not answered correctly, the student gets the marks only for the first question.
- If both the questions are answered correctly, the student gets the sum of the marks of the two questions.
The following table shows the probability of correctly answering a question and the marks of the question respectively.

Assuming that the student always wants to maximize her expected marks in the examination, in which order should she attempt the questions and what is the expected marks for that order (assume that the questions are independent)?

A.
First QuesA and then QuesB. Expected marks 14.
B.
First QuesB and then QuesA. Expected marks 14.
C.
First QuesB and then QuesA. Expected marks 22.
D.
First QuesA and then QuesB. Expected marks 16.

Correct Answer:

Step-by-Step Solution

Insight: To maximize expected marks, calculate the expected payoff for each possible ordering (A then B, or B then A) using a decision tree.

Exam route:

Order A then B: $0.8 \times (0.5 \times 30 + 0.5 \times 10) = 0.8 \times 20 = 16$ .

Order B then A: $0.5 \times (0.8 \times 30 + 0.2 \times 20) = 0.5 \times 28 = 14$ .

Max is 16, so attempt QuesA first, then QuesB.

Learning route:

Define strategies: Strategy 1 is A then B. Strategy 2 is B then A.
Calculate expected marks for Strategy 1 (A then B):

Probability of getting A wrong: $0.2$ . Marks = $0$ .
Probability of getting A right, B wrong: $0.8 \times 0.5 = 0.4$ . Marks = $10$ .
Probability of getting both right: $0.8 \times 0.5 = 0.4$ . Marks = $10 + 20 = 30$ .
Expected marks = $0.2 (0) + 0.4 (10) + 0.4 (30) = 0 + 4 + 12 = 16$ .

Calculate expected marks for Strategy 2 (B then A):

Probability of getting B wrong: $0.5$ . Marks = $0$ .
Probability of getting B right, A wrong: $0.5 \times 0.2 = 0.1$ . Marks = $20$ .
Probability of getting both right: $0.5 \times 0.8 = 0.4$ . Marks = $20 + 10 = 30$ .
Expected marks = $0.5 (0) + 0.1 (20) + 0.4 (30) = 0 + 2 + 12 = 14$ .

Compare: $16 > 14$ , so attempt QuesA first to maximize expected marks.

Question 5 · Engineering Mathematics · 2021_Set2 NAT

For a given biased coin, the probability that the outcome of a toss is a head is 0.4. This coin is tossed 1,000 times. Let $X$ denote the random variable whose value is the number of times that head appeared in these 1,000 tosses. The standard deviation of $X$ (rounded to 2 decimal places) is __________.

Correct Answer:

15.49

Step-by-Step Solution

Insight: The number of heads in $n$ independent tosses of a biased coin follows a Binomial distribution $B (n, p)$ .

Exam route: Identify $n = 1000$ , $p = 0.4$ , $q = 0.6$ . Variance $= n pq = 1000 \times 0.4 \times 0.6 = 240$ . Standard deviation $= 240 \approx 15.49$ .

Learning route:

Recognize the distribution: $X \sim Binomial (1000, 0.4)$ .
Recall the variance formula for a Binomial random variable: $Var (X) = n p (1 - p)$ .
Calculate the variance: $Var (X) = 1000 \times 0.4 \times 0.6 = 240$ .
Standard deviation is the square root of variance: $σ_{X} = 240 \approx 15.4919$ .
Round to two decimal places as requested: $15.49$ .

Question 6 · Engineering Mathematics · 2021_Set1 MCQ

Consider the two statements.

S_{1} : There exist random variables X and Y such that

(E [(X - E (X)) (Y - E (Y))])^{2} > Var [X] Var [Y]

S_{2} : For all random variables X and Y,

Cov [X, Y] = E [∣ X - E [X] ∣ ∣ Y - E [Y] ∣]

Which one of the following choices is correct?

A.
Both $S_{1}$ and $S_{2}$ are true.
B.
$S_{1}$ is true, but $S_{2}$ is false.
C.
$S_{1}$ is false, but $S_{2}$ is true.
D.
Both $S_{1}$ and $S_{2}$ are false.

Correct Answer:

Step-by-Step Solution

Key idea: this is a statement-truth question on covariance. Recognise that the inner expectation in $S_{1}$ is exactly $Cov [X, Y]$ , and that the RHS of $S_{2}$ is an expectation of a non-negative quantity.

Why these methods apply:

$S_{1}$ is a claim about an inequality involving a square of a covariance-like term and a product of variances. The standard tool is the Cauchy–Schwarz inequality for random variables.
$S_{2}$ equates $Cov [X, Y]$ with an expectation involving absolute values. The standard tool is to check the sign behaviour: covariance can be negative, but an expectation of a product of absolute values cannot.

Step 1 — decode $S_{1}$ :

By definition

$E [(X - E X) (Y - E Y)] = Cov [X, Y] .$

So $S_{1}$ is the claim: there exist $X, Y$ with $(Cov [X, Y])^{2} > Var [X] Var [Y]$ .

Step 2 — apply Cauchy–Schwarz:

For any two random variables $U, V$ with finite second moments

$(E [U V])^{2} \leq E [U^{2}] E [V^{2}] .$

Choose $U = X - E X$ and $V = Y - E Y$ . Then $E [U^{2}] = Var [X]$ , $E [V^{2}] = Var [Y]$ , and $E [U V] = Cov [X, Y]$ . Hence

$(Cov [X, Y])^{2} \leq Var [X] Var [Y]$

for every pair $X, Y$ . Strict " $>$ " is impossible, so $S_{1}$ is FALSE.

Step 3 — decode $S_{2}$ :

$Cov [X, Y] = E [(X - E X) (Y - E Y)]$ , which can be negative.

The RHS of $S_{2}$ is $E [∣ X - E X ∣ ∣ Y - E Y ∣]$ , an expectation of a non-negative random variable, so it is always $\geq 0$ .

Therefore $S_{2}$ cannot hold for all $X, Y$ .

Step 4 — concrete counterexample for $S_{2}$ :

Let $X$ be any non-constant zero-mean random variable and set $Y = - X$ . Then $E X = E Y = 0$

$Cov [X, Y] = E [X (- X)] = - Var [X] < 0$

while

$E [∣ X ∣ ∣ Y ∣] = E [X^{2}] = Var [X] > 0.$

The two sides disagree, so $S_{2}$ is FALSE.

Conclusion: both $S_{1}$ and $S_{2}$ are false.

Common trap: reading $S_{1}$ as "some cleverly chosen $X, Y$ beat the inequality" — but Cauchy–Schwarz is universal, no choice of $X, Y$ can violate it. Another trap: forgetting that absolute values destroy the sign information in $S_{2}$ .

Verification: plug $U = X - E X, V = Y - E Y$ into Cauchy–Schwarz — the inequality $(Cov [X, Y])^{2} \leq Var [X] Var [Y]$ is recovered exactly, confirming $S_{1}$ is false; the $Y = - X$ counterexample confirms $S_{2}$ is false.

Answer: option (D) — Both $S_{1}$ and $S_{2}$ are false.

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Random Variables, Expectation, Variance and Covariance PYQs for GATE CS

Chapter Roadmap: Probability and Statistics

Chapter Journey: Probability and Statistics

1. Random Variables and Distributions

2. Expectation and Moments (Current Topic)

3. Standard Distributions

4. Limit Theorems

Core Idea: Geometric Expectation via Integration

The Core Idea

General Method

Key Steps

Random Variables, Expectation, Variance and Covariance: Solved Questions with Step-by-Step Explanations (6 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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