The value of $correlation(X, Y)$ is __________ . (Answer in integer)
We need to find the correlation between random variables $X$ and $Y$, denoted as $Corr(X, Y)$.
Understanding Random Variable Distributions
- $X$ follows a $Uniform(-1, 1)$ distribution.
- The conditional distribution of $Y$ given $X=x$ is $Uniform(x^2 - 0.1, x^2 + 0.1)$.
Calculating Expected Values
First, let's find the expected value of $X$ and $Y$ given $X=x$.
- For $X \sim Uniform(a, b)$, the expected value is $E[X] = \frac{a+b}{2}$.
So, $E[X] = \frac{-1 + 1}{2} = 0$. - For $Y|X=x \sim Uniform(x^2 - 0.1, x^2 + 0.1)$, the conditional expectation is: $E[Y|X=x] = \frac{(x^2 - 0.1) + (x^2 + 0.1)}{2} = \frac{2x^2}{2} = x^2$.
Calculating Overall Expected Value E[Y]
Using the law of total expectation, $E[Y] = E[E[Y|X]]$:
$E[Y] = E[X^2]$
For $X \sim Uniform(a, b)$, $E[X^2] = \frac{a^2 + ab + b^2}{3}$.
Here, $a = -1$ and $b = 1$.
$E[X^2] = \frac{(-1)^2 + (-1)(1) + (1)^2}{3} = \frac{1 - 1 + 1}{3} = \frac{1}{3}$.
Therefore, $E[Y] = \frac{1}{3}$.
Calculating Covariance Cov(X, Y)
The formula for covariance is $Cov(X, Y) = E[XY] - E[X]E[Y]$.
We need $E[XY]$. Using the law of total expectation again:
$E[XY] = E[E[XY|X]] = E[X \cdot E[Y|X]]$
Substitute $E[Y|X] = X^2$:
$E[XY] = E[X \cdot X^2] = E[X^3]$.
For $X \sim Uniform(-1, 1)$, $E[X^k] = \frac{1}{1 - (-1)} \int_{-1}^{1} x^k dx$.
Calculate $E[X^3]$:
$E[X^3] = \frac{1}{2} \int_{-1}^{1} x^3 dx = \frac{1}{2} \left[ \frac{x^4}{4} \right]_{-1}^{1} = \frac{1}{2} \left( \frac{1^4}{4} - \frac{(-1)^4}{4} \right) = \frac{1}{2} \left( \frac{1}{4} - \frac{1}{4} \right) = 0$.
So, $E[XY] = 0$.
Now, calculate the covariance:
$Cov(X, Y) = E[XY] - E[X]E[Y] = 0 - (0) \cdot (\frac{1}{3}) = 0$.
Calculating Correlation Corr(X, Y)
The correlation is calculated as $Corr(X, Y) = \frac{Cov(X, Y)}{\sqrt{Var(X) Var(Y)}}$.
Since $Cov(X, Y) = 0$, and the variances $Var(X)$ and $Var(Y)$ are finite and non-zero (specifically $Var(X) = \frac{(1-(-1))^2}{12} = \frac{4}{12} = \frac{1}{3}$), the correlation is:
$Corr(X, Y) = \frac{0}{\sqrt{Var(X) Var(Y)}} = 0$.
Final Answer
The value of $correlation(X, Y)$ is 0.
Important Questions from Correlation Analysis
If X ∼ N (0, 1) and Y = X2 then the correlation coefficient r (X, Y) is
Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): If the securities with less than perfect negative correlation between their price movements are combined, portfolio risk can be reduced significantly.
Reason (R): The term with negative correlation has the effect of reducing the computed value of total portfolio risk, given other terms that are positive.
In the light of the above statements, choose the most appropriate answer from the options given below:
The two-regression equation of variable \(\rm{x}\) and \(\rm{y}\) are
\(\rm{y = 0.8x + 9.8}\) and \(\rm{x = 10.2 + 0.6y}\)
The coefficient of correlation between \(\rm{x}\) and \(\rm{y}\) is
- In the variance-covariance matrix given below, which one of the following pairs of elements has the highest correlation?
$$ \begin{bmatrix} & P & Q & R & S \\ P & 0 & 0 & 0 & 0 \\ Q & 0 & 3.73 & 1.85 & 3.23 \\ R & 0 & 1.85 & 0.93 & 1.61 \\ S & 0 & 3.23 & 1.61 & 2.79 \end{bmatrix} $$ - Consider 62 pair of observations $(x_i, y_i)$, i = 1, …., 62, obtained from the distribution of $(X, Y)$. The Pearson correlation coefficient ($R$) of these observations is 0.27.
Which one of the following statements is CORRECT for a significance level $\alpha = 0.05$ and $t_{0.025,60} = 2.0$?
Hint: Test statistic $t = \frac{R\sqrt{n - 2}}{\sqrt{1 - R^2}}$
