GATE DA

Random variables

Random variables, expectation and variance, including conditional expectation

A random variable turns the outcome of an experiment into a number, and its distribution (PMF, PDF or CDF) tells you how probability is spread over those numbers. Expectation, variance, covariance and conditional expectation summarise that distribution. Problems on this topic are solved by applying linearity of expectation, the variance rules, and the laws of total expectation and total variance correctly.

8 previous-year questions · 8 solved · 5 practice questions

Random variables, PMF, PDF and CDF

A random variable $X$ assigns a real number to each outcome of an experiment. It is discrete if it takes countably many values and is described by a probability mass function (PMF). It is continuous if it has a probability density function (PDF); then $P (X = x) = 0$ for every single point $x$ .

PMF CONDITIONS (DISCRETE)

p_{X} (x) = P (X = x) \geq 0, x \sum p_{X} (x) = 1

Use the second condition to find an unknown constant in a given PMF.

PDF CONDITIONS (CONTINUOUS)

f_{X} (x) \geq 0, \int_{- \infty}^{\infty} f_{X} (x) d x = 1, P (a \leq X \leq b) = \int_{a}^{b} f_{X} (x) d x

A density can exceed 1 at a point; only its integral is a probability.

CDF

F_{X} (x) = P (X \leq x) = {\sum_{t \leq x} p_{X} (t) \int_{- \infty}^{x} f_{X} (t) d t discrete continuous

The CDF is non-decreasing, right-continuous, with limits 0 at $- \infty$ and 1 at $+ \infty$ . For any random variable, $P (a < X \leq b) = F_{X} (b) - F_{X} (a)$ .

RECOVERING PMF AND PDF FROM THE CDF

f_{X} (x) = \frac{d F _{X} ( x )}{d x}, p_{X} (x) = F_{X} (x) - F_{X} (x^{-})

For a continuous variable differentiate the CDF where it is differentiable; for a discrete variable the PMF is the size of the jump at $x$ .

EXAM TIP

For a continuous variable $P (a < X < b) = P (a \leq X \leq b)$ , since single points carry no probability. For a discrete variable the end points matter, so read $<$ and $\leq$ carefully.

Expectation and variance

EXPECTATION

E [X] = x \sum x p_{X} (x) or E [X] = \int_{- \infty}^{\infty} x f_{X} (x) d x

Defined when the sum or integral converges absolutely.

EXPECTATION OF A FUNCTION (LOTUS)

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

No need to find the distribution of $g (X)$ . In general $E [g (X)] \neq = g (E [X])$ ; they are equal when $g$ is linear, $g (x) = a x + b$ .

LINEARITY OF EXPECTATION

E [a X + bY + c] = a E [X] + b E [Y] + c, E [i = 1 \sum n X_{i}] = i = 1 \sum n E [X_{i}]

True whenever the expectations exist, with or without independence.

VARIANCE

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

Standard deviation $SD = Var (X)$ . Since $Var (X) \geq 0$ , $E [X^{2}] \geq (E [X])^{2}$ , and $Var (X) = 0$ only when $X$ is constant with probability 1. Convention: $Var (X)$ here is the population (distribution) variance. For a list of $n$ data values, the population variance divides by $n$ and the sample variance divides by $n - 1$ .

VARIANCE UNDER A LINEAR CHANGE

Var (a X + b) = a^{2} Var (X)

A shift $b$ does not change the variance; a scale $a$ multiplies it by $a^{2}$ (the sign of $a$ is lost).

MOMENTS

μ_{k}^{'} = E [X^{k}], μ_{k} = E [(X - E [X])^{k}]

The $k$ -th raw moment and $k$ -th central moment; $μ_{2} = Var (X)$ .

EXPECTATION OF A NON-NEGATIVE INTEGER VARIABLE

E [X] = k = 1 \sum \infty P (X \geq k)

Valid for $X$ taking values in ${0, 1, 2, \dots}$ ; often quicker than summing $k p_{X} (k)$ .

INDICATOR VARIABLES

I_{A} = {10 A occurs otherwise, E [I_{A}] = P (A), Var (I_{A}) = P (A) (1 - P (A))

To find the expected number of events $A_{1}, \dots, A_{n}$ that occur, write the count as $\sum I_{A_{i}}$ ; by linearity its mean is $\sum P (A_{i})$ , with no independence needed.

Mean and variance of standard distributions

Distribution	Parameters	Mean	Variance
Bernoulli	$p$	$p$	$p (1 - p)$
Binomial	$n, p$	$n p$	$n p (1 - p)$
Geometric (trials up to and including the first success, support $1, 2, \dots$ )	$p$	$1/ p$	$(1 - p) / p^{2}$
Geometric (failures before the first success, support $0, 1, 2, \dots$ )	$p$	$(1 - p) / p$	$(1 - p) / p^{2}$
Poisson	$λ$	$λ$	$λ$
Discrete uniform on ${1, \dots, n}$	$n$	$(n + 1) /2$	$(n^{2} - 1) /12$
Continuous uniform on $(a, b)$	$a, b$	$(a + b) /2$	$(b - a)^{2} /12$
Exponential (rate $λ$ )	$λ$	$1/ λ$	$1/ λ^{2}$
Normal	$μ, σ^{2}$	$μ$	$σ^{2}$

WORKED EXAMPLE

A continuous random variable has PDF $f (x) = k x^{2}$ for $0 \leq x \leq 2$ and $0$ elsewhere. Find $k$ , $E [X]$ and $Var (X)$ .

Solution

Normalise: $\int_{0}^{2} k x^{2} d x = \frac{8 k}{3} = 1$ , so $k = \frac{3}{8}$ .
$E [X] = \int_{0}^{2} x \cdot \frac{3}{8} x^{2} d x = \frac{3}{8} \cdot \frac{2 ^{4}}{4} = \frac{3}{2}$ .
$E [X^{2}] = \int_{0}^{2} \frac{3}{8} x^{4} d x = \frac{3}{8} \cdot \frac{32}{5} = \frac{12}{5}$ .
$Var (X) = E [X^{2}] - (E [X])^{2} = 2.4 - 2.25 = 0.15$ .

Answer: $k = 3/8$ , $E [X] = 1.5$ , $Var (X) = 0.15$ .

Joint distributions, independence, covariance

For two random variables the joint PMF is $p_{X, Y} (x, y) = P (X = x, Y = y)$ and the joint PDF $f_{X, Y} (x, y)$ satisfies $P ((X, Y) \in A) = \iint_{A} f_{X, Y} d x d y$ . Both are non-negative and total 1.

MARGINALS

p_{X} (x) = y \sum p_{X, Y} (x, y), f_{X} (x) = \int_{- \infty}^{\infty} f_{X, Y} (x, y) d y

Sum (or integrate) out the other variable over its whole range. Mind the limits when the support is not a rectangle.

INDEPENDENCE

p_{X, Y} (x, y) = p_{X} (x) p_{Y} (y) for all x, y (f_{X, Y} (x, y) = f_{X} (x) f_{Y} (y))

For a PMF the product rule must hold at every pair $(x, y)$ , so one failing pair shows dependence. For a PDF it must hold everywhere apart from sets of zero probability. If the set where the joint PMF or PDF is positive is not a product set $A \times B$ (for example the triangle $0 < x < y < 1$ ), then $X$ and $Y$ are dependent.

EXPECTATION OF A FUNCTION OF TWO VARIABLES

E [g (X, Y)] = x \sum y \sum g (x, y) p_{X, Y} (x, y) or \iint g (x, y) f_{X, Y} (x, y) d x d y

Used for $E [X Y]$ and similar.

COVARIANCE

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

$Cov (X, X) = Var (X)$ ; $Cov (a X + b, c Y + d) = a c Cov (X, Y)$ .

CORRELATION COEFFICIENT

ρ_{X Y} = \frac{Cov ( X , Y )}{Var ( X ) Var ( Y )}, - 1 \leq ρ_{X Y} \leq 1

Defined when both variances are positive. $∣ ρ ∣ = 1$ exactly when $Y = a X + b$ with probability 1 for some constants $a \neq = 0$ and $b$ ; then $ρ = 1$ if $a > 0$ and $ρ = - 1$ if $a < 0$ .

VARIANCE OF A SUM

Var (a X + bY) = a^{2} Var (X) + b^{2} Var (Y) + 2 ab Cov (X, Y)

For uncorrelated (in particular independent) variables the covariance term vanishes, so $Var (X \pm Y) = Var (X) + Var (Y)$ . Note the plus sign, even for the difference.

VARIANCE OF A SUM OF N VARIABLES

Var (i = 1 \sum n X_{i}) = i = 1 \sum n Var (X_{i}) + 2 i < j \sum Cov (X_{i}, X_{j})

For independent identically distributed $X_{i}$ with variance $σ^{2}$ , the sample mean has $Var (\overset{ˉ}{X}) = σ^{2} / n$ .

PRODUCT OF INDEPENDENT VARIABLES

X, Y independent \Rightarrow E [X Y] = E [X] E [Y], Cov (X, Y) = 0

The converse is false: zero covariance (uncorrelated) does not imply independence.

EXAM TIP

Classic trap: take $X$ symmetric about 0 (for example uniform on ${- 1, 0, 1}$ ) and $Y = X^{2}$ . Then $E [X] = 0$ and $E [X Y] = E [X^{3}] = 0$ , so $Cov (X, Y) = 0$ , yet $Y$ is completely determined by $X$ , so they are dependent.

Conditional distributions and conditional expectation

CONDITIONAL PMF AND PDF

p_{X ∣ Y} (x ∣ y) = \frac{p _{X, Y} ( x , y )}{p _{Y} ( y )}, f_{X ∣ Y} (x ∣ y) = \frac{f _{X, Y} ( x , y )}{f _{Y} ( y )}

Requires $p_{Y} (y) > 0$ (or $f_{Y} (y) > 0$ ). If $X, Y$ are independent the conditional equals the marginal.

CONDITIONAL EXPECTATION

E [X ∣ Y = y] = x \sum x p_{X ∣ Y} (x ∣ y) or \int x f_{X ∣ Y} (x ∣ y) d x

For each fixed $y$ this is a number; $E [X ∣ Y]$ is the random variable obtained by replacing $y$ with $Y$ .

CONDITIONAL VARIANCE

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

Also a random variable, a function of $Y$ .

LAW OF TOTAL EXPECTATION

E [X] = E [E [X ∣ Y]] = y \sum E [X ∣ Y = y] P (Y = y)

Replace the sum by an integral against $f_{Y}$ when $Y$ is continuous. For events $A_{1}, A_{2}, \dots$ that partition the sample space, $E [X] = \sum_{i} E [X ∣ A_{i}] P (A_{i})$ .

LAW OF TOTAL VARIANCE

Var (X) = E [Var (X ∣ Y)] + Var (E [X ∣ Y])

Average within-group variance plus variance of the group means.

USEFUL PROPERTIES

E [a X + b Z ∣ Y] = a E [X ∣ Y] + b E [Z ∣ Y], E [g (Y) X ∣ Y] = g (Y) E [X ∣ Y], E [X ∣ Y] = E [X] if X, Y independent

Once $Y$ is given, any function of $Y$ behaves like a constant. The last property does not work in reverse: $E [X ∣ Y] = E [X]$ does not imply independence.

RANDOM SUM

S = i = 1 \sum N X_{i} : E [S] = E [N] μ, Var (S) = E [N] σ^{2} + Var (N) μ^{2}

Here $X_{1}, X_{2}, \dots$ are independent and identically distributed with mean $μ$ and variance $σ^{2}$ , and $N$ is a non-negative integer random variable independent of the $X_{i}$ . It follows from the two total laws with $E [S ∣ N] = N μ$ and $Var (S ∣ N) = N σ^{2}$ .

Procedure for a mixture or two-stage problem

Identify the conditioning variable $Y$ (the first stage, or the group label).
Write down $E [X ∣ Y]$ and $Var (X ∣ Y)$ as functions of $Y$ .
Find $E [X]$ by averaging $E [X ∣ Y]$ over the distribution of $Y$ .
Find $Var (X)$ as the mean of $Var (X ∣ Y)$ plus the variance of $E [X ∣ Y]$ .

WORKED EXAMPLE

A component is of type A with probability $0.6$ and of type B with probability $0.4$ . A type A lifetime has mean $10$ and variance $4$ ; a type B lifetime has mean $20$ and variance $25$ . Find the mean and variance of the lifetime $X$ of a randomly chosen component.

Solution

Let $Y$ be the type. $E [X] = 0.6 (10) + 0.4 (20) = 14$ .
$E [Var (X ∣ Y)] = 0.6 (4) + 0.4 (25) = 12.4$ .
$E [(E [X ∣ Y])^{2}] = 0.6 (100) + 0.4 (400) = 220$ , so $Var (E [X ∣ Y]) = 220 - 1 4^{2} = 24$ .
$Var (X) = 12.4 + 24 = 36.4$ .

Answer: $E [X] = 14$ and $Var (X) = 36.4$ .

EXAM TIP

A mixture's variance is at least as large as the average of the component variances, and equal only when every component has the same mean. Forgetting the $Var (E [X ∣ Y])$ term is the commonest slip.

Markov's and Chebyshev's inequalities

MARKOV'S INEQUALITY

P (X \geq a) \leq \frac{E [ X ]}{a}

For $X \geq 0$ and $a > 0$ . Needs only the mean.

CHEBYSHEV'S INEQUALITY

P (∣ X - μ ∣ \geq k σ) \leq \frac{1}{k ^{2}}, equivalently P (∣ X - μ ∣ \geq t) \leq \frac{σ ^{2}}{t ^{2}}

Holds for any distribution with mean $μ$ and finite variance $σ^{2}$ ; $k, t > 0$ . Complementary form: $P (∣ X - μ ∣ < k σ) \geq 1 - \frac{1}{k ^{2}}$ .

Both give upper bounds that are valid for every distribution of the stated kind, so they can be far from the actual probability. A bound is not the probability itself. A one-sided question such as $P (X - μ \geq t)$ is also bounded by $σ^{2} / t^{2}$ , since that event lies inside $∣ X - μ ∣ \geq t$ .

EXAM TIP

Markov needs $X \geq 0$ . Chebyshev is Markov applied to $(X - μ)^{2}$ with threshold $t^{2}$ . To bound a probability that a sample mean is far from $μ$ , use $Var (\overset{ˉ}{X}) = σ^{2} / n$ for independent identically distributed samples.

PREVIOUS-YEAR QUESTIONS

8 questions from the DA papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE DA 2026 · GATE DA 2025 · GATE DA 2024

GATE 2026 DA Q.44

GATE 2026 · DAQ.44 · 2 marks · MCQ

Let $X$ and $Y$ be two independent random variables. $X$ follows $B er n o u l l i (p = 0.3)$ distribution and $Y$ follows $N or ma l (μ = 0, σ^{2} = 100)$ distribution.

Which of the following options is the variance of $(2 X - 1) Y$ ?

(A)
100
(B)
90
(C)
49
(D)
21

Answer and solution

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GATE 2026 DA Q.54

GATE 2026 · DAQ.54 · 2 marks · MSQ

Let $X$ be a discrete valued random variable with cumulative distribution function $F (x)$ .

Which of the following statements is/are correct?

(A)
$F (x)$ is always a positive function.
(B)
$F (x)$ is a non-decreasing function.
(C)
$F (x)$ has jump discontinuity.
(D)
$F (x)$ is a left continuous function.

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GATE 2025 DA Q.11

GATE 2025 · DAQ.11 · 1 mark · MCQ

Suppose $X$ and $Y$ are random variables. The conditional expectation of $X$ given $Y$ is denoted by $E [X ∣ Y]$ . Then $E [E [X ∣ Y]]$ equals

(A)
$E [X ∣ Y]$
(B)
$\frac{E [ X ]}{E [ Y ]}$
(C)
$E [X]$
(D)
$E [Y]$

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GATE 2025 DA Q.20

GATE 2025 · DAQ.20 · 1 mark · MCQ

Let $X = a Z + b$ , where $Z$ is a standard normal random variable, and $a$ , $b$ are two unknown constants. It is given that $E [X] = 1$ , $E [(X - E [X]) Z] = - 2$ , $E [(X - E [X])^{2}] = 4$ , where $E [X]$ denotes the expectation of random variable $X$ . The values of $a, b$ are:

(A)
$a = - 2, b = 1$
(B)
$a = 2, b = - 1$
(C)
$a = - 2, b = - 1$
(D)
$a = 1, b = 1$

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GATE 2025 DA Q.36

GATE 2025 · DAQ.36 · 2 marks · MCQ

Let $Y = Z^{2}$ , $Z = \frac{X - μ}{σ}$ , where $X$ is a normal random variable with mean $μ$ and variance $σ^{2}$ . The variance of $Y$ is

(A)
1
(B)
2
(C)
3
(D)
4

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GATE 2025 DA Q.54

GATE 2025 · DAQ.54 · 2 marks · MSQ

Consider a coin-toss experiment where the probability of head showing up is $p$ . In the $i^{th}$ coin toss, let $X_{i} = 1$ if head appears, and $X_{i} = 0$ if tail appears. Consider $p = \frac{1}{n} \sum_{i = 1}^{n} X_{i}$ where $n$ is the total number of independent coin tosses. Which of the following statements is/are correct?

(A)
$E [p] = p$
(B)
$E [p] = \frac{p}{n}$
(C)
As $n$ increases, variance of $p$ decreases
(D)
Variance of $p$ does not depend on $n$

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GATE 2024 DA Q.36

GATE 2024 · DAQ.36 · 2 marks · MCQ

A fair six-sided die (with faces numbered 1, 2, 3, 4, 5, 6) is repeatedly thrown independently.

What is the expected number of times the die is thrown until two consecutive throws of even numbers are seen?

(A)
2
(B)
4
(C)
6
(D)
8

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GATE 2024 DA Q.59

GATE 2024 · DAQ.59 · 2 marks · NAT

Consider a joint probability density function of two random variables $X$ and $Y$

f_{X, Y} (x, y) = {2 x y, 0, 0 < x < 2, 0 < y < x otherwise

Then, $E [Y ∣ X = 1.5]$ is ______.

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PRACTICE

5 questions to test yourself

PRACTICE 11 mark · MCQ

A discrete random variable $X$ has PMF $P (X = k) = c k$ for $k = 1, 2, 3, 4$ and $0$ otherwise. The value of $E [X]$ is

(A)
$2.5$
(B)
$3$
(C)
$3.5$
(D)
$4$

Answer and solution

Answer: (B)

Normalise: $c (1 + 2 + 3 + 4) = 10 c = 1$ , so $c = 0.1$ .
$E [X] = \sum k \cdot (0.1 k) = 0.1 (1 + 4 + 9 + 16) = 0.1 \times 30 = 3$ .

PRACTICE 22 marks · NAT

Random variables $X$ and $Y$ have $Var (X) = 4$ , $Var (Y) = 9$ and correlation coefficient $ρ_{X Y} = 0.5$ . The value of $Var (2 X - Y)$ is ______.

Answer and solution

Answer: $13$ (accepted range $\pm 0.01$ )

$Cov (X, Y) = ρ 4 \cdot 9 = 0.5 \times 6 = 3$ .
$Var (2 X - Y) = 4 Var (X) + Var (Y) - 2 (2) (1) Cov (X, Y)$ .
$= 16 + 9 - 12 = 13$ .

PRACTICE 32 marks · MSQ

The joint PMF of $X, Y \in {0, 1}$ is $P (0, 0) = 0.2$ , $P (0, 1) = 0.3$ , $P (1, 0) = 0.3$ , $P (1, 1) = 0.2$ . Which of the following statements is/are TRUE?

(A)
$X$ and $Y$ are independent
(B)
$Cov (X, Y) = - 0.05$
(C)
$E [Y ∣ X = 0] = 0.6$
(D)
$Var (X + Y) = 0.4$

Answer and solution

Answer: (B), (C), (D)

Marginals: $P (X = 0) = P (X = 1) = 0.5$ and likewise for $Y$ , so $E [X] = E [Y] = 0.5$ and $Var (X) = Var (Y) = 0.25$ .
A is false: $P (0, 0) = 0.2 \neq = 0.5 \times 0.5 = 0.25$ .
B: $E [X Y] = P (1, 1) = 0.2$ , so $Cov = 0.2 - 0.25 = - 0.05$ . True.
C: $E [Y ∣ X = 0] = P (Y = 1 ∣ X = 0) = 0.3/0.5 = 0.6$ . True.
D: $Var (X + Y) = 0.25 + 0.25 + 2 (- 0.05) = 0.4$ . True.

PRACTICE 42 marks · NAT

A fair six-faced die is rolled once, giving $N$ . A fair coin is then tossed $N$ times and $X$ is the number of heads. The value of $Var (X)$ , correct to two decimal places, is ______.

Answer and solution

Answer: $1.6042$ (accepted range $\pm 0.01$ )

Given $N = n$ , $X$ is Binomial $(n, 1/2)$ : $E [X ∣ N] = N /2$ , $Var (X ∣ N) = N /4$ .
$E [N] = 3.5$ and $Var (N) = 35/12$ for a fair die.
$E [Var (X ∣ N)] = 3.5/4 = 0.875 = 42/48$ .
$Var (E [X ∣ N]) = Var (N) /4 = 35/48 \approx 0.7292$ .
$Var (X) = 42/48 + 35/48 = 77/48 \approx 1.60$ .

PRACTICE 51 mark · MCQ

A continuous random variable $X$ has PDF $f (x) = 2 x$ for $0 \leq x \leq 1$ and $0$ elsewhere. The value of $Var (X)$ is

(A)
$1/18$
(B)
$1/12$
(C)
$1/9$
(D)
$1/6$

Answer and solution

Answer: (A)

$E [X] = \int_{0}^{1} x \cdot 2 x d x = 2/3$ .
$E [X^{2}] = \int_{0}^{1} x^{2} \cdot 2 x d x = 1/2$ .
$Var (X) = 1/2 - 4/9 = 1/18$ .

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE DA

conditional expectation and variance … random variables, discrete random variables and probability mass functions

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