Question:medium
Let \(X\), \(N\), \(Y\) and \(Z\) be random variables. The variables \(X\) and \(N\) are independent of each other. \(X\) is uniformly distributed between \(-1\) and \(1\); \(N\) follows Normal distribution with zero mean and unity variance.
\(Y\) and \(Z\) are defined as \(Y=X+N\) and \(Z=X^2+N\).
Which of the following pairs represents the values of correlation between \(X\) and \(Y\), and that between \(X\) and \(Z\)?
Let \(X\), \(N\), \(Y\) and \(Z\) be random variables. The variables \(X\) and \(N\) are independent of each other. \(X\) is uniformly distributed between \(-1\) and \(1\); \(N\) follows Normal distribution with zero mean and unity variance.
\(Y\) and \(Z\) are defined as \(Y=X+N\) and \(Z=X^2+N\).
Which of the following pairs represents the values of correlation between \(X\) and \(Y\), and that between \(X\) and \(Z\)?
\(Y\) and \(Z\) are defined as \(Y=X+N\) and \(Z=X^2+N\).
Which of the following pairs represents the values of correlation between \(X\) and \(Y\), and that between \(X\) and \(Z\)?
Show Hint
Use Corr(U,V) = E[UV] - E[U]E[V]; the odd moment E[X^3] of a symmetric uniform variable is zero.
Updated On: Jul 20, 2026
- \(\dfrac{1}{3}\) and \(0\)
- \(\dfrac{1}{3}\) and \(\dfrac{1}{9}\)
- \(\dfrac{1}{3}\) and \(\dfrac{1}{3}\)
- \(1\) and \(0\)
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Write correlation as a covariance.
For random variables $U,V$, take correlation to mean $E[UV]-E[U]E[V]$. Both $X$ and $N$ have zero mean here, which simplifies things a lot.
Step 2: Collect the moments of X we need.
$X$ is uniform on $[-1,1]$. Its mean is $0$. Its variance is
\[ \text{Var}(X)=\frac{(1-(-1))^2}{12}=\frac{1}{3} \]
Because the density is an even function, all odd powers of $X$ average to zero, so $E[X^3]=0$ as well.
Step 3: Handle the noise term N.
$N$ has zero mean and is independent of $X$, so any product like $E[XN]$ or $E[X^2N]$ factors and gives zero.
Step 4: Correlate X with Y.
\[ E[XY]=E[X(X+N)]=E[X^2]+E[XN]=\frac{1}{3}+0=\frac{1}{3} \]
Since $E[X]E[Y]=0$, the correlation between $X$ and $Y$ is $\frac{1}{3}$.
Step 5: Correlate X with Z.
\[ E[XZ]=E[X(X^2+N)]=E[X^3]+E[XN]=0+0=0 \]
So the correlation between $X$ and $Z$ is $0$.
Step 6: State the pair.
\[ \boxed{\frac{1}{3}\text{ and }0} \]
For random variables $U,V$, take correlation to mean $E[UV]-E[U]E[V]$. Both $X$ and $N$ have zero mean here, which simplifies things a lot.
Step 2: Collect the moments of X we need.
$X$ is uniform on $[-1,1]$. Its mean is $0$. Its variance is
\[ \text{Var}(X)=\frac{(1-(-1))^2}{12}=\frac{1}{3} \]
Because the density is an even function, all odd powers of $X$ average to zero, so $E[X^3]=0$ as well.
Step 3: Handle the noise term N.
$N$ has zero mean and is independent of $X$, so any product like $E[XN]$ or $E[X^2N]$ factors and gives zero.
Step 4: Correlate X with Y.
\[ E[XY]=E[X(X+N)]=E[X^2]+E[XN]=\frac{1}{3}+0=\frac{1}{3} \]
Since $E[X]E[Y]=0$, the correlation between $X$ and $Y$ is $\frac{1}{3}$.
Step 5: Correlate X with Z.
\[ E[XZ]=E[X(X^2+N)]=E[X^3]+E[XN]=0+0=0 \]
So the correlation between $X$ and $Z$ is $0$.
Step 6: State the pair.
\[ \boxed{\frac{1}{3}\text{ and }0} \]
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