Variance Calculation for Random Variables
We are given two random variables, $X$ and $Y$, with the following statistical properties:
- Means: $E[X] = 0$, $E[Y] = 0$
- Variances: $\text{Var}(X) = 1$, $\text{Var}(Y) = 1$
- Correlation Coefficient: $\rho_{XY} = \frac{1}{3}$
The objective is to find the value of $\text{Var}(X + 3Y)$.
Variance Formula for Linear Combinations
The general formula for the variance of a linear combination of two random variables, $aX + bY$, is:
$ \text{Var}(aX + bY) = a^2 \text{Var}(X) + b^2 \text{Var}(Y) + 2ab \text{Cov}(X, Y) $
In this problem, $a=1$ and $b=3$. We first need to find the covariance $\text{Cov}(X, Y)$.
Covariance Calculation
The covariance is related to the correlation coefficient by the formula:
$ \text{Cov}(X, Y) = \rho_{XY} \sqrt{\text{Var}(X)} \sqrt{\text{Var}(Y)} $
Substituting the given values:
$ \text{Cov}(X, Y) = \frac{1}{3} \sqrt{1} \sqrt{1} = \frac{1}{3} $
Final Variance Calculation
Now, we apply the variance formula with $a=1$, $b=3$, $\text{Var}(X)=1$, $\text{Var}(Y)=1$, and $\text{Cov}(X, Y)=\frac{1}{3}$:
$ \text{Var}(X + 3Y) = (1)^2 \text{Var}(X) + (3)^2 \text{Var}(Y) + 2(1)(3) \text{Cov}(X, Y) $
$ \text{Var}(X + 3Y) = 1 \cdot (1) + 9 \cdot (1) + 6 \cdot \left(\frac{1}{3}\right) $
$ \text{Var}(X + 3Y) = 1 + 9 + 2 $
$ \text{Var}(X + 3Y) = 12 $
The value of $\text{Var}(X + 3Y)$ is 12.
Important Questions from Variance
Consider a distribution with the following probability density function $$f(x) = \begin{cases} 0.5, & 0 < x < 2 \\ 0.0, & Otherwise \end{cases}$$ Given that the mean of the above probability distribution is 1, the variance (rounded off to two decimal places) is _______________.
- Let $X$ and $Y$ be two independent random variables. $X$ follows $Bernoulli(p = 0.3)$ distribution and $Y$ follows $Normal(\mu = 0, \sigma^2 = 100)$ distribution.
Which of the following options is the variance of $(2X - 1)Y$? - For a given data set $\{x_1, x_2, \ldots, x_n\}$, where $n = 100$, it is known that
$$ \frac{1}{2000} \sum_{i=1}^{n} \sum_{j=1}^{n} (x_i - x_j)^2 = 99 $$
Let us denote $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$.
The value of $\frac{1}{99} \sum_{i=1}^{n} (x_i - \bar{x})^2$ is __________ . (Answer in integer) - A random variable $X$ has the sample space $\{0,1\}$. The probability $P(X = 0) = 1/4$ and $P(X = 1) = 3/4$.
What is the variance of the random variable?
Hint: $\text{Mean } (\mu) = \sum_{i=1}^{n} x_i p(x_i) ; \text{Variance } (\sigma^2) = \sum_{i=1}^{n} (x_i - \mu)^2 p(x_i)$ The unbiased sample variance for the set of numbers: $S = \{40,45,50,55,60\}$ is_____. (write answer with one decimal place)
