Random Variables, Expectation, Variance and Covariance Practice Questions for GATE CS

Solve 104+ Random Variables, Expectation, Variance and Covariance practice 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 (10 Problems)

Question 1 · Engineering Mathematics MCQ

Let $X$ be a discrete random variable taking values in the set ${0, 1, 2, 3, 4}$ . What is the maximum possible value of the expected value $E [X]$ ?

A.
2
B.
3
C.
3.5
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: The expected value is bounded by the minimum and maximum possible values of the random variable.

Step 1: Recall that $E [X] = \sum x_{i} P (X = x_{i})$ is a weighted average of the possible values.

Step 2: A weighted average is maximized when all weight is on the largest value.

Step 3: The maximum value in ${0, 1, 2, 3, 4}$ is 4. This occurs when $P (X = 4) = 1$ and all other probabilities are 0.

Step 4: In this case, $E [X] = 4 \times 1 = 4$ .

Answer: 4

Common mistake: Thinking $E [X]$ is always the middle value (2) or the average of the range (3.5).

Question 2 · Engineering Mathematics MCQ

Consider the following three games:

\begin{itemize}

\item Game A: Win 10 with probability 0.5, lose 2 with probability 0.5

\item Game B: Win 8 with probability 0.6, lose 4 with probability 0.4

\item Game C: Win 6 with probability 0.7, lose 3 with probability 0.3

\end{itemize}

Match each game with its expected payoff and identify the optimal game to play.

A.
A: 4, B: 3.2, C: 3.3; Optimal: A
B.
A: 4, B: 3.2, C: 3.3; Optimal: B
C.
A: 6, B: 4.8, C: 4.2; Optimal: A
D.
A: 4, B: 3.3, C: 3.2; Optimal: C

Correct Answer:

Step-by-Step Solution

Key idea: This is a decision problem where we compute the expected payoff for each option and choose the maximum.

Step 1: Compute $E [payoff]$ for each game. "Lose" means negative payoff.

Step 2: Game A:

$E [A] = 10 (0.5) + (- 2) (0.5) = 5 - 1 = 4$

Step 3: Game B:

$E [B] = 8 (0.6) + (- 4) (0.4) = 4.8 - 1.6 = 3.2$

Step 4: Game C:

$E [C] = 6 (0.7) + (- 3) (0.3) = 4.2 - 0.9 = 3.3$

Step 5: Compare: $E [A] = 4 > E [C] = 3.3 > E [B] = 3.2$ .

Answer: A: 4, B: 3.2, C: 3.3; Optimal: A

Common mistake: Misreading "lose" as "win" or swapping B and C values.

Question 3 · Engineering Mathematics MCQ

Let $X$ be a random variable that takes values in the interval $[0, 1]$ with $E [X] = 0.5$ . The maximum possible value of $Var (X)$ is:

A.
$0.25$
B.
$0.50$
C.
$0.75$
D.
$1.00$

Correct Answer:

Step-by-Step Solution

Key idea: This is an optimization problem using the computational formula for variance and the constraint that $X \in [0, 1]$ .

Step 1: Recall the computational formula:

$Var (X) = E [X^{2}] - (E [X])^{2} = E [X^{2}] - 0.25$

Step 2: To maximize $Var (X)$ , we need to maximize $E [X^{2}]$ .

Step 3: Since $X \in [0, 1]$ , we have $X^{2} \leq X$ (because $X \leq 1$ implies $X^{2} \leq X$ ).

Step 4: Therefore, $E [X^{2}] \leq E [X] = 0.5$ .

Step 5: The maximum value of $E [X^{2}]$ is $0.5$ , achieved when $X$ is a Bernoulli random variable with $P (X = 1) = 0.5$ and $P (X = 0) = 0.5$ .

Step 6: Maximum variance:

$Var (X) = 0.5 - 0.25 = 0.25$

Answer: A

Common trap: A student might think $E [X^{2}]$ can be as large as $1$ (since $X^{2} \leq 1$ ), giving $Var (X) = 1 - 0.25 = 0.75$ (option C). However, this ignores the constraint that $E [X] = 0.5$ , which limits how large $E [X^{2}]$ can be.

Question 4 · Engineering Mathematics MCQ

Let $X$ and $Y$ be any random variables. Which of the following statements is ALWAYS true?

A.
$Var (X + Y) = Var (X) + Var (Y)$
B.
$Var (2 X) = 2 Var (X)$
C.
$Var (X + 5) = Var (X)$
D.
$Var (X - Y) = Var (X) - Var (Y)$

Correct Answer:

Step-by-Step Solution

Key idea: This is a statement truth problem testing the properties of variance, specifically the effect of constants and the requirement of independence.

Step 1: Analyze each option:

Option A: $Var (X + Y) = Var (X) + Var (Y)$

This is ONLY true if $X$ and $Y$ are independent (or uncorrelated)
In general: $Var (X + Y) = Var (X) + Var (Y) + 2 Cov (X, Y)$
NOT always true ✗

Option B: $Var (2 X) = 2 Var (X)$

The correct property is: $Var (a X) = a^{2} Var (X)$
So $Var (2 X) = 4 Var (X)$ , not $2 Var (X)$
NOT true ✗

Option C: $Var (X + 5) = Var (X)$

Adding a constant shifts the distribution but doesn't change the spread
Property: $Var (X + b) = Var (X)$ for any constant $b$
ALWAYS true ✓

Option D: $Var (X - Y) = Var (X) - Var (Y)$

The correct formula is: $Var (X - Y) = Var (X) + Var (Y) - 2 Cov (X, Y)$
This is NOT $Var (X) - Var (Y)$
NOT true ✗

Answer: C

Common trap: A student might choose option A, forgetting that $Var (X + Y) = Var (X) + Var (Y)$ only holds when $X$ and $Y$ are independent. The problem states "any random variables," so we cannot assume independence.

Question 5 · Engineering Mathematics MCQ

Let $X$ and $Y$ be discrete random variables. Which of the following statements is ALWAYS true?

A.
$E [X^{2}] = (E [X])^{2}$
B.
$E [X + Y] = E [X] + E [Y]$ only if $X$ and $Y$ are independent
C.
$E [a X + b] = a E [X] + b$ for any constants $a, b$
D.
$E [X]$ must be a possible value of $X$

Correct Answer:

Step-by-Step Solution

Key idea: This is a statement truth problem testing the properties of discrete expectation, specifically linearity and common misconceptions.

Step 1: Analyze each option:

Option A: $E [X^{2}] = (E [X])^{2}$

This is FALSE in general
The correct relationship is: $Var (X) = E [X^{2}] - (E [X])^{2}$
So $E [X^{2}] = (E [X])^{2} + Var (X) \geq (E [X])^{2}$
Equality holds only when $Var (X) = 0$ (i.e., $X$ is constant)
NOT always true ✗

Option B: $E [X + Y] = E [X] + E [Y]$ only if $X$ and $Y$ are independent

This is FALSE
Linearity of expectation holds regardless of independence
$E [X + Y] = E [X] + E [Y]$ is ALWAYS true
NOT always true ✗

Option C: $E [a X + b] = a E [X] + b$ for any constants $a, b$

This is the linearity property of expectation
It holds for any random variable $X$ and any constants $a, b$
ALWAYS true ✓

Option D: $E [X]$ must be a possible value of $X$

This is FALSE
Example: Fair die roll has $E [X] = 3.5$ , but 3.5 is not a possible outcome
NOT always true ✗

Answer: C

Common trap: A student might choose option B, thinking that linearity requires independence. However, linearity of expectation is one of the most powerful properties precisely because it holds regardless of dependence.

Question 6 · Engineering Mathematics MCQ

Let $X$ be a random variable with $E [X] = 3$ and $X \in [0, 6]$ . The maximum possible value of $Var (X)$ is:

A.
$6$
B.
$9$
C.
$12$
D.
$18$

Correct Answer:

Step-by-Step Solution

Key idea: This is an optimization problem using the computational formula for variance and the constraint that $X \in [0, 6]$ .

Step 1: Recall the computational formula:

$Var (X) = E [X^{2}] - (E [X])^{2} = E [X^{2}] - 9$

Step 2: To maximize $Var (X)$ , we need to maximize $E [X^{2}]$ .

Step 3: For a random variable $X \in [a, b]$ with fixed mean $μ$ , the maximum variance is achieved by a Bernoulli-like distribution that puts all probability mass at the endpoints $a$ and $b$ .

Step 4: The maximum variance formula for $X \in [a, b]$ with $E [X] = μ$ is:

$Var_{m a x} = (μ - a) (b - μ)$

Step 5: Substitute the values:

$Var_{m a x} = (3 - 0) (6 - 3) = 3 \times 3 = 9$

Answer: B

Common trap: A student might think the maximum variance occurs when $X$ is uniformly distributed on $[0, 6]$ , giving $Var (X) = \frac{( 6 - 0 ) ^{2}}{12} = 3$ (option A is not this, but this is a common mistake). Or they might forget the constraint $E [X] = 3$ and think the maximum is when $X$ takes values $0$ and $6$ with equal probability, giving $E [X] = 3$ and $Var (X) = 9$ , which is correct but they might not realize this is the maximum.

Question 7 · Engineering Mathematics MCQ

A discrete random variable $X$ takes values $1, 2, 3$ with probabilities $P (X = 1) = \frac{1}{2}$ , $P (X = 2) = \frac{1}{4}$ , and $P (X = 3) = \frac{1}{4}$ . Using the Law of the Unconscious Statistician, what is $E [X^{2}]$ ?

A.
3.06
B.
3.50
C.
3.75
D.
4.25

Correct Answer:

Step-by-Step Solution

Key idea: LOTUS (Law of the Unconscious Statistician) allows computing $E [g (X)]$ directly without finding the distribution of $g (X)$ .

Step 1: Recall LOTUS: $E [g (X)] = \sum g (x_{i}) P (X = x_{i})$ .

Step 2: Here $g (X) = X^{2}$ , so we apply the square to each value:

$E [X^{2}] = 1^{2} \times \frac{1}{2} + 2^{2} \times \frac{1}{4} + 3^{2} \times \frac{1}{4}$

Step 3: Calculate:

$E [X^{2}] = 1 \times 0.5 + 4 \times 0.25 + 9 \times 0.25$

$E [X^{2}] = 0.5 + 1.0 + 2.25 = 3.75$

Answer: 3.75

Common mistake: Computing $E [X]^{2}$ instead of $E [X^{2}]$ . First find $E [X] = 1 (0.5) + 2 (0.25) + 3 (0.25) = 1.75$ , then $1.7 5^{2} = 3.0625 \approx 3.06$ .

Question 8 · Engineering Mathematics NAT

Let $X$ and $Y$ be random variables, not necessarily independent, that take real values in the interval $[0, 1]$ . If $E [X] = 0.6$ and $E [Y] = 0.5$ , how many of the following values can be the value of $E [X Y]$ ?

$1.$ $0.05$

$2.$ $0.10$

$3.$ $0.30$

$4.$ $0.50$

Correct Answer:

3.00

Step-by-Step Solution

Key idea: This is a casework problem using both the upper and lower bounds for the expectation of a product of bounded random variables.

Step 1: Compute the upper bound:

$E [X Y] \leq min (E [X], E [Y]) = min (0.6, 0.5) = 0.5$

Step 2: Compute the lower bound:

$E [X Y] \geq max (0, E [X] + E [Y] - 1) = max (0, 0.6 + 0.5 - 1) = max (0, 0.1) = 0.1$

Step 3: The range of possible values for $E [X Y]$ is $[0.1, 0.5]$ .

Step 4: Check each value:

$0.05 < 0.1$ : NOT in range ✗
$0.10 = 0.1$ : In range (lower boundary) ✓
$0.30 \in [0.1, 0.5]$ : In range ✓
$0.50 = 0.5$ : In range (upper boundary) ✓

Step 5: Count the valid values: $3$

Answer: 3

Common trap: A student might compute the lower bound as $1 - 0.6 - 0.5 = - 0.1$ (sign error), then $max (0, - 0.1) = 0$ , giving range $[0, 0.5]$ . This would incorrectly include $0.05$ , yielding a count of $4$ .

Question 9 · Engineering Mathematics NAT

Let $X$ be a random variable. How many of the following statements are ALWAYS true?

$1.$ $Var (X) \geq 0$

$2.$ $Var (X) = 0$ implies $X$ is a constant

$3.$ $Var (X + 5) = Var (X) + 25$

$4.$ $Var (2 X) = 2 Var (X)$

Correct Answer:

2.00

Step-by-Step Solution

Key idea: This is a casework problem testing the properties of variance.

Step 1: Analyze each statement:

Statement 1: $Var (X) \geq 0$

Variance is the expected value of squared deviations
Squared values are always non-negative
Expectation of non-negative values is non-negative
TRUE ✓

Statement 2: $Var (X) = 0$ implies $X$ is a constant

If $Var (X) = 0$ , then $E [(X - μ)^{2}] = 0$
Since $(X - μ)^{2} \geq 0$ , the only way the expectation is 0 is if $(X - μ)^{2} = 0$ almost surely
This means $X = μ$ almost surely, so $X$ is a constant
TRUE ✓

Statement 3: $Var (X + 5) = Var (X) + 25$

Adding a constant shifts the distribution but doesn't change spread
Property: $Var (X + b) = Var (X)$ for any constant $b$
So $Var (X + 5) = Var (X)$ , not $Var (X) + 25$
FALSE ✗

Statement 4: $Var (2 X) = 2 Var (X)$

Scaling property: $Var (a X) = a^{2} Var (X)$
So $Var (2 X) = 2^{2} Var (X) = 4 Var (X)$ , not $2 Var (X)$
FALSE ✗

Step 2: Count the true statements: 2

Answer: 2

Common trap: A student might make a sign error and think $Var (X + 5) = Var (X) + 25$ by incorrectly applying the scaling property to addition, or think $Var (2 X) = 2 Var (X)$ by forgetting to square the constant.

Question 10 · Engineering Mathematics MCQ

A student claims that for any fixed point $c \in (0, 1)$ , when the unit interval $(0, 1)$ is divided at a point chosen uniformly at random, the expected length of the subinterval containing $c$ is always greater than or equal to $1$ . The actual minimum value of $E (c) = c - c^{2} + \frac{1}{2}$ that contradicts this statement is:

A.
$r a c 12$
B.
$r a c 34$
C.
$1$
D.
$r a c 14$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct optimization problem using the formula $E (c) = c - c^{2} + \frac{1}{2}$ .

Step 1: To find the minimum, take the derivative and set it to zero:

$\frac{d E}{d c} = 1 - 2 c = 0 ⟹ c = \frac{1}{2}$

Step 2: Substitute $c = \frac{1}{2}$ into $E (c)$ :

$E (\frac{1}{2}) = \frac{1}{2} - (\frac{1}{2})^{2} + \frac{1}{2} = \frac{1}{2} - \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$

Step 3: Verify this is a minimum by checking the second derivative:

$\frac{d ^{2} E}{d c ^{2}} = - 2 < 0$

Wait, this indicates a maximum, not a minimum! Let me reconsider.

Actually, $E (c) = c - c^{2} + \frac{1}{2}$ is a downward-opening parabola (coefficient of $c^{2}$ is negative), so $c = \frac{1}{2}$ gives the MAXIMUM value, not the minimum.

The minimum occurs at the boundary. Since $c \in (0, 1)$ , we check the limits:

As $c \to 0^{+}$ : $E (c) \to 0 - 0 + \frac{1}{2} = \frac{1}{2}$
As $c \to 1^{-}$ : $E (c) \to 1 - 1 + \frac{1}{2} = \frac{1}{2}$

So the minimum value is $\frac{1}{2}$ , achieved as $c$ approaches $0$ or $1$ .

Answer: A

Common trap: A student might overcount by thinking the expected length should account for both subintervals, getting $E = 1$ (option C). But the formula $E (c) = c - c^{2} + \frac{1}{2}$ already gives the correct expected length of the single subinterval containing $c$ .

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Random Variables, Expectation, Variance and Covariance Practice Questions 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 (10 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

Step-by-Step Solution

Step-by-Step Solution

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Step-by-Step Solution

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