GATE DA 2026 | Question: 53

Question

Let $X$ be a random variable that follows Uniform $(-1,1)$ distribution. The conditional distribution of the random variable $Y$ given $X=x$ is the Uniform( $x^{2}-0.1, x^{2}+0.1$ ) distribution.

The value of correlation $(X, Y)$ is $\_\_\_\_$. (Answer in integer)

Answer 1

Given:

$$X \sim \text{Uniform}(-1,1) \implies f_{X}=\frac{1}{1-(-1)}=\frac{1}{2}$$

$$X\mid Y \sim \text{Uniform}(x^2-0.1, x^2+0.1)\implies f_{X \mid Y} = \frac{1}{x^2+0.1-(x^2-0.1)}=\frac{1}{0.2}=5$$

We know that $\text{Corr}(\rho)=\dfrac{\text{Cov(X,Y)}}{\sigma_X \times\sigma_Y}$

So calculating Covariance:

$$\text{Cov}(X,Y)=\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)$$

 

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Notice we need Joint Density ($f_{XY}$) to calculate $\mathbb{E}(XY)$

We know,  $\boxed{\text{Conditional} = \dfrac{\text{Joint}}{\text{Marginal}}}$

Substitute above to get Joint $f_{XY}=\dfrac{5}2$

Start calculating terms one by one:

$$\mathbb{E}(XY)= \int_{-1}^{1}\int_{x^2-0.1}^{x^2+0.1} xy \times f_{XY} \space dy \space dx = \frac{5}{2} \int_{-1}^{1}x\int_{x^2-0.1}^{x^2+0.1}y \space dy \space dx$$

$$\frac{5}{2}\times\frac{1}{10} \int_{-1}^{1}x \times x^2 \space dx  = \boxed{0}$$

Now for :

$$\mathbb{E}(X) = \int_{-1}^{1} x \times f_{X} \space dx = \frac{1}{2} \int_{-1}^{1}x \space dx = \boxed{0}$$

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Place the obtained values in the above Covariance equation.

$\because \text{Cov}(X,Y) = 0$, Correlation is also 0.

---

Hence `0` is accurate.

Answer 2

Given:

\[
X \sim \text{Uniform}(-1,1)
\]

\[
Y \mid X=x \sim \text{Uniform}(x^2 - 0.1,\; x^2 + 0.1)
\]

Step 1: Compute conditional expectation

For a uniform distribution on \([a,b]\), the mean is

\[
\frac{a+b}{2}.
\]

Thus,

\[
E[Y \mid X=x]
=
\frac{(x^2 - 0.1) + (x^2 + 0.1)}{2}
=
x^2.
\]

Hence,

\[
E[Y \mid X] = X^2.
\]

 

Step 2: Compute covariance

We use

\[
\mathrm{Cov}(X,Y) = E[XY] - E[X]E[Y].
\]

To compute \(E[XY]\), we use the Law of Total Expectation
(Tower Property):

\[
E[XY] = E\big[E[XY \mid X]\big].
\]

Since we are conditioning on X, the variable X becomes constant inside the conditional expectation.

Therefore,

\[
E[XY \mid X] = X E[Y \mid X].
\]

Thus,

\[
E[XY] = E\big[X E[Y \mid X]\big].
\]

Since \(E[Y \mid X] = X^2\),

\[
E[XY] = E[X \cdot X^2] = E[X^3].
\]

Step 3: Use symmetry

Because

\[
X \sim \text{Uniform}(-1,1),
\]

the distribution is symmetric about 0.

Hence all odd moments are zero:

\[
E[X^3] = 0.
\]

Alternative Method to Compute \(E[X^3]\):

Since the density of \(X\) is

\[
f(x) = \frac{1}{2}, \quad -1 \le x \le 1,
\]

we compute directly:

\[
E[X^3]
=
\int_{-1}^{1} x^3 \cdot \frac{1}{2} \, dx
=
\frac{1}{2}
\int_{-1}^{1} x^3 \, dx.
\]

Because \(x^3\) is an odd function and the interval \([-1,1]\) is symmetric,

\[
\int_{-1}^{1} x^3 \, dx = 0.
\]

Therefore,

\[
E[X^3] = 0.
\]

Also,

\[
E[X] = 0.
\]

Therefore,

\[
\mathrm{Cov}(X,Y) = 0.
\]

Step 4: Compute correlation

\[
\mathrm{Corr}(X,Y)
=
\frac{\mathrm{Cov}(X,Y)}
{\sqrt{\mathrm{Var}(X)\mathrm{Var}(Y)}}.
\]

Since covariance is zero,

\[
\mathrm{Corr}(X,Y) = 0.
\]

Final Answer:

\[
\boxed{0}
\]

Answer 3

$X \sim \text{Uniform}(-1, 1), \quad Y|X=x \sim U(x^2 - 0.1, x^2 + 0.1)$

$E[X] = \frac{a + b}{2} = \frac{1 - 1}{2} = 0$

$E[Y|X=x] = \frac{(x^2 - 0.1) + (x^2 + 0.1)}{2} = x^2$

$\begin{aligned} E[Y] &= E[E[Y|X]] = E[X^2]  \quad (\because \text{since } E[Y|X] = X^2) \end{aligned}$

$\begin{aligned} \text{cov}(X, Y) &= E[XY] - \overbrace{E[X]}^{0}E[Y] \\ &= E[X X^2] \\&= E[X^3] \quad (\because \text{As } E[Y] = E[X^2]) \end{aligned}$

$E[X^3] = \int_{-1}^{1} x^3 \frac{1}{b - a} dx =\int_{-1}^{1} x^3 \frac{1}{1 - (-1)} dx = \frac{1}{2} \left[ \frac{x^4}{4} \right]_{-1}^{1} = 0$

$\begin{aligned} \text{So, cov}(X, Y) = 0 \end{aligned}$

$\text{Thus, correlation}(X,Y) = \frac{\text{cov}(X,Y)}{\sigma_X \cdot \sigma_Y} = 0$

Answer 4

You can solve this within seconds. Let me show you The Trick :

Correlation  = Measures the Strength of a Linear Relation between 2 elements.

1  => Denotes 1 as X increases , Y also Increases.

-1 => As X increases, Y decreases.

0 => NO straight line relationship.

 

 

Comments

alright, but i didnt understand how and why odd powers would positively correlated and even powers would be uncorrelated, with mere inspection.
All i know is that if $X \sim \text{Uniform(-1,1)}$, then $X^2 \sim\text{UnknownDistribution}$.

u must have seen such question earlier, i want to know how you came to the conclusion that odd powers are '+' correlated whereas even powers are uncorrelated?

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This is a Trick I, normally use  to identify stuffs like these in seconds... That's why I have shared this trick with the students !!!!  Although I am a regular in GO, but,  normally I don't have time to study different question papers. Someday, I would Love to do so !!! Enjoy The Trick and Try to apply it  wherever it suits !!!!

God Bless you

 

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Nice trick brother :-)