If X ∼ N (0, 1) and Y = X2 then the correlation coefficient r (X, Y) is
zero
Calculating the Correlation Coefficient between a Standard Normal Variable and its Square
The question asks for the correlation coefficient between a standard normal random variable X and a variable Y, where Y is defined as the square of X, i.e., $Y = X^2$. X follows a standard normal distribution, denoted as X ∼ N (0, 1).
A standard normal distribution N(0, 1) has the following properties:
- Mean E[X] = 0
- Variance Var[X] = 1
The probability density function of X is symmetric about the mean (0). This symmetry is important when calculating expected values of odd powers of X.
The correlation coefficient between two random variables X and Y is defined as:
$$ \rho (X, Y) = \frac{Cov(X, Y)}{\sqrt{Var(X)Var(Y)}} $$Where $Cov(X, Y)$ is the covariance between X and Y, given by $Cov(X, Y) = E[XY] - E[X]E[Y]$.
Step-by-Step Calculation of Correlation Coefficient
To find the correlation coefficient r(X, Y), we need to calculate the following values:
- Expected value of X, E[X]
- Expected value of Y, E[Y]
- Expected value of XY, E[XY]
- Variance of X, Var(X)
- Variance of Y, Var(Y)
- Covariance of X and Y, Cov(X, Y)
1. Expected Value of X (E[X])
Since X ∼ N(0, 1), the mean is 0.
$E[X] = 0$
2. Expected Value of Y (E[Y])
Y is defined as $Y = X^2$. The expected value of Y is E[Y] = E[X²].
For any random variable X, the variance is given by $Var[X] = E[X^2] - (E[X])^2$.
Rearranging this formula, we get $E[X^2] = Var[X] + (E[X])^2$.
Using the properties of N(0, 1): Var[X] = 1 and E[X] = 0.
$E[Y] = E[X^2] = Var[X] + (E[X])^2 = 1 + (0)^2 = 1 + 0 = 1$
3. Expected Value of XY (E[XY])
Since $Y = X^2$, XY = $X \times X^2 = X^3$. We need to find E[X³].
For a standard normal distribution (which is symmetric about 0), all odd moments about the mean are zero. The third moment about the mean is E[(X - E[X])³]. Since E[X] = 0, this is E[X³].
Therefore, $E[X^3] = 0$ for X ∼ N(0, 1).
$E[XY] = E[X^3] = 0$
4. Variance of X (Var(X))
Given that X ∼ N(0, 1), the variance is 1.
$Var(X) = 1$
5. Variance of Y (Var(Y))
To find Var(Y), we use the formula $Var(Y) = E[Y^2] - (E[Y])^2$.
We already found E[Y] = 1. Now we need E[Y²].
$E[Y^2] = E[(X^2)^2] = E[X^4]$
For a standard normal distribution N(0, 1), the fourth moment E[X⁴] is 3.
$E[Y^2] = E[X^4] = 3$
Now calculate Var(Y):
$Var(Y) = E[Y^2] - (E[Y])^2 = 3 - (1)^2 = 3 - 1 = 2$
6. Covariance of X and Y (Cov(X, Y))
Using the formula $Cov(X, Y) = E[XY] - E[X]E[Y]$.
We have E[XY] = 0, E[X] = 0, and E[Y] = 1.
$Cov(X, Y) = 0 - (0)(1) = 0 - 0 = 0$
7. Correlation Coefficient r(X, Y)
Finally, substitute the calculated values into the correlation coefficient formula:
$$ r(X, Y) = \frac{Cov(X, Y)}{\sqrt{Var(X)Var(Y)}} $$ $$ r(X, Y) = \frac{0}{\sqrt{1 \times 2}} = \frac{0}{\sqrt{2}} = 0 $$Summary of Calculations
| Quantity | Value for X ∼ N(0, 1) | Calculation/Reason |
|---|---|---|
| E[X] | $0$ | Mean of N(0, 1) |
| E[Y] = E[X²] | $1$ | Var(X) + (E[X])² = 1 + 0² |
| E[XY] = E[X³] | $0$ | Odd moment of symmetric distribution about 0 |
| Var(X) | $1$ | Variance of N(0, 1) |
| Var(Y) = E[X⁴] - (E[X²])² | $2$ | 3 - 1² (E[X⁴]=3 for N(0,1)) |
| Cov(X, Y) | $0$ | E[XY] - E[X]E[Y] = 0 - 0*1 |
| r(X, Y) | $0$ | Cov(X, Y) / $\sqrt{Var(X)Var(Y))}$ = 0 / $\sqrt{1*2}$ |
The correlation coefficient r(X, Y) is 0.
A correlation coefficient of zero means there is no linear relationship between X and Y. It does not imply that X and Y are independent, especially when a non-linear relationship like Y = X² exists.
Important Questions from Correlation Analysis
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:
X, Y and Z are three uncorrelated variables having variances \(\sigma_x^2, \sigma_y^2 \:and\:\sigma_z^2\) respectively, then the correlation between X + Y and Y + Z is:
Calculate the correlation coefficient between the following values :
x: 3, 5, 1, 7, 5
y: 4, 3, 0, 8, 2
Consider two exponentially distributed random variables X and Y, both having a mean of 0.50. Let Z = X + Y and r be the correlation coefficient between X and Y. If the variance of Z equals 0, then the value of r is _______ (round off to 2 decimal places).
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
