GATE DA 2026 | Question: 34

Question

Let $X$ and $Y$ be two independent random variables. $X$ follows $\operatorname{Bernoulli}(p=0.3)$ distribution and $Y$ follows $\operatorname{Normal}\left(\mu=0, \sigma^{2}=100\right)$ distribution.

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

• A. $100$
• B. $90$
• C. $49$
• D. $21$

Comments

Mistake may happen just like me : we can’t write  Varinace(2XY-Y) = Variance(2XY) + Variance(Y) = 2^2*Variance(XY)+Variance(Y) because 2XY and Y are not independent

---

Important and Useful Note:

If X and Y are two independent random variables and f, g are two functions, then f(X) and g(Y) are also independent. Independence means knowing information about X doesn’t tell anything about Y. Applying separate transformations to each random variable cannot create dependence between them.

Answer 1

$X \sim \text{Bernoulli}(0.3), \quad
Y \sim N(0,100)$

$X$ and $Y$ are independent.

We have to find:
\[
\operatorname{Var}((2X-1)Y)
\]

Method 1: Using expansion of variance

Since $X$ and $Y$ are independent, we have
\[
\mathbb{E}[XY] = \mathbb{E}[X]\mathbb{E}[Y],
\]
because independence implies
\[
\operatorname{Cov}(X,Y)=0.
\]

Now,

\[
\operatorname{Var}((2X-1)Y)
=
\mathbb{E}[(2X-1)^2 Y^2]
-
\left(\mathbb{E}[(2X-1)Y]\right)^2
\]

Since $X$ and $Y$ are independent,

\[
\mathbb{E}[(2X-1)Y]
=
\mathbb{E}[2X-1]\mathbb{E}[Y]
\]

But $\mathbb{E}[Y]=0$, hence

\[
\mathbb{E}[(2X-1)Y]=0
\]

Therefore,

\[
\operatorname{Var}((2X-1)Y)
=
\mathbb{E}[(2X-1)^2]\mathbb{E}[Y^2]
\]

Now,

\[
\mathbb{E}[Y^2]
=
\operatorname{Var}(Y) + (\mathbb{E}[Y])^2
=
100
\]

Next compute:

\[
(2X-1)^2 = 4X^2 - 4X + 1
\]

So,

\[
\mathbb{E}[(2X-1)^2]
=
4\mathbb{E}[X^2]
-
4\mathbb{E}[X]
+
1
\]

Using

\[
\mathbb{E}[X^2] = \operatorname{Var}(X) + (\mathbb{E}[X])^2
\]

we get

\[
=
4(\operatorname{Var}(X)+(\mathbb{E}[X])^2)
-
4\mathbb{E}[X]
+
1
\]

For $X \sim \text{Bernoulli}(p)$:

\[
\mathbb{E}[X]=p, 
\quad 
\operatorname{Var}(X)=p(1-p)
\]

Thus,

\[
=
4p(1-p) + 4p^2 - 4p + 1
\]

\[
=
4p - 4p^2 + 4p^2 - 4p + 1
\]

\[
= 1
\]

Hence,

\[
\operatorname{Var}((2X-1)Y)
=
1 \times 100
=
100
\]

 

Method 2: Using Bernoulli property  $X^2 = X$

 

If $X \sim \text{Bernoulli}(p)$, then $X \in \{0,1\}$.

Hence, for any integer $n \ge 1$,
\[
X^n = X.
\]

Now,

Since $X$ is Bernoulli, $X \in \{0,1\}$ and hence

\[
X^2 = X
\]

Then,

\[
(2X-1)^2
=
4X^2 - 4X + 1
=
4X - 4X + 1
=
1
\]

So,

\[
\mathbb{E}[(2X-1)^2] = 1
\]

Therefore,

\[
\operatorname{Var}((2X-1)Y)
=
\mathbb{E}[(2X-1)^2]\mathbb{E}[Y^2]
=
1 \times 100
=
100
\]

\bigskip
\[
\boxed{\operatorname{Var}((2X-1)Y)=100}
\]