Question

Let $X$ be a discrete random variable with probability mass function $P(X=x_i)$ where $x \in \{x_1, x_2, \dots, x_n\}$. If $\sum_{i=1}^{n} (x_i-2)^2 P(X=x_i) = 28$ and $\sum_{i=1}^{n} (x_i+2)^2 P(X=x_i) = 12$. Then variance of $X$ is?

• A. $20$
• B. $12$
• C. $8$
• D. $40$

Hint:

• Use $E[g(X)] = \sum g(x_i) P(X=x_i)$.
• Given $E[(X-2)^2] = 28 \Rightarrow E[X^2] - 4E[X] + 4 = 28 \Rightarrow E[X^2] - 4E[X] = 24$.
• Given $E[(X+2)^2] = 12 \Rightarrow E[X^2] + 4E[X] + 4 = 12 \Rightarrow E[X^2] + 4E[X] = 8$.
• Solve these two equations: Adding them gives $2E[X^2] = 32 \Rightarrow E[X^2] = 16$. Subtracting the first from the second gives $8E[X] = -16 \Rightarrow E[X] = -2$.
• Variance $Var(X) = E[X^2] - (E[X])^2 = 16 - (-2)^2 = 16 - 4 = 12$.

Answer 1

The Correct Option is B

We are given a discrete random variable $X$ with the probability mass function $P(X = x_i)$ where $x \in \{x_1, x_2, \dots, x_n\}$, and the following information:

• $\sum_{i=1}^{n} (x_i - 2)^2 P(X = x_i) = 28$
• $\sum_{i=1}^{n} (x_i + 2)^2 P(X = x_i) = 12$

We are tasked with finding the variance of $X$.

Step 1: Express the given sums in terms of expected values

The $n^{th}$ term $T_n$ of a progression can be found by subtracting the sum of the first $n-1$ terms from the sum of the first $n$ terms:

We can express the two given sums as expectations:

First sum:

\[ \sum_{i=1}^{n} (x_i - 2)^2 P(X = x_i) \]

Expanding the square:

\[ (x_i - 2)^2 = x_i^2 - 4x_i + 4 \] \[ \sum_{i=1}^{n} (x_i - 2)^2 P(X = x_i) = \sum_{i=1}^{n} \left( x_i^2 - 4x_i + 4 \right) P(X = x_i) \] \[ = \sum_{i=1}^{n} x_i^2 P(X = x_i) - 4 \sum_{i=1}^{n} x_i P(X = x_i) + 4 \sum_{i=1}^{n} P(X = x_i) \] Using properties of expectations: \[ \sum_{i=1}^{n} P(X = x_i) = 1 \quad \text{(since the total probability sums to 1)} \] Thus, the equation becomes: \[ \mathbb{E}[X^2] - 4\mathbb{E}[X] + 4 = 28 \]

Second sum:

\[ \sum_{i=1}^{n} (x_i + 2)^2 P(X = x_i) \] Expanding the square: \[ (x_i + 2)^2 = x_i^2 + 4x_i + 4 \] \[ \sum_{i=1}^{n} (x_i + 2)^2 P(X = x_i) = \sum_{i=1}^{n} \left( x_i^2 + 4x_i + 4 \right) P(X = x_i) \] \[ = \sum_{i=1}^{n} x_i^2 P(X = x_i) + 4 \sum_{i=1}^{n} x_i P(X = x_i) + 4 \sum_{i=1}^{n} P(X = x_i) \] Again, using properties of expectations: \[ \mathbb{E}[X^2] + 4\mathbb{E}[X] + 4 = 12 \]

Step 2: Solving the system of equations

We now have the following system of equations: 1. $\mathbb{E}[X^2] - 4\mathbb{E}[X] + 4 = 28$ 2. $\mathbb{E}[X^2] + 4\mathbb{E}[X] + 4 = 12$ Let $\mu = \mathbb{E}[X]$ and $\mu_2 = \mathbb{E}[X^2]$. Substituting into the system, we have: 1. $\mu_2 - 4\mu + 4 = 28$ 2. $\mu_2 + 4\mu + 4 = 12$ Subtract the second equation from the first: \[ (\mu_2 - 4\mu + 4) - (\mu_2 + 4\mu + 4) = 28 - 12 \] Simplifying: \[ \mu_2 - 4\mu + 4 - \mu_2 - 4\mu - 4 = 16 \] \[ -8\mu = 16 \] \[ \mu = -2 \] Now substitute $\mu = -2$ into one of the original equations, say the first one: \[ \mu_2 - 4(-2) + 4 = 28 \] \[ \mu_2 + 8 + 4 = 28 \] \[ \mu_2 + 12 = 28 \] \[ \mu_2 = 16 \]

Step 3: Finding the variance

The variance of $X$, denoted $\text{Var}(X)$, is given by: \[ \text{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 \] Substituting the values of $\mathbb{E}[X^2] = 16$ and $\mathbb{E}[X] = -2$: \[ \text{Var}(X) = 16 - (-2)^2 = 16 - 4 = 12 \]

Final Answer:

The variance of $X$ is 12.