If a random variable X is defined such that E[(X - 1)²] = 10, and E[(X - 2)²] = 6. Find the mean of X.

Understand the Problem

The question is asking to find the mean of a random variable X given two expected values related to its deviations from specific numbers. We will apply concepts from statistics to derive the mean based on the provided equations.

Answer

The mean of $X$ is $\mu = 3.5$.

Steps to Solve

Understand the equations for expected values

The given equations are: $$ E[(X - 1)^2] = 10 $$ $$ E[(X - 2)^2] = 6 $$

These express the expected values of the squared deviations of $X$ from 1 and 2.

Expand the equations

We can expand the squared terms:

For the first equation: $$ E[(X - 1)^2] = E[X^2 - 2X + 1] = E[X^2] - 2E[X] + 1 = 10 $$
For the second equation: $$ E[(X - 2)^2] = E[X^2 - 4X + 4] = E[X^2] - 4E[X] + 4 = 6 $$

Set up a system of equations

From the expansions, we can create two equations (denote $E[X]$ as $\mu$):

$$ E[X^2] - 2\mu + 1 = 10 \quad \Rightarrow \quad E[X^2] = 2\mu + 9 \tag{1} $$
$$ E[X^2] - 4\mu + 4 = 6 \quad \Rightarrow \quad E[X^2] = 4\mu + 2 \tag{2} $$
Equate the equations for (E[X^2])

Set the right-hand sides of (1) and (2) equal to each other: $$ 2\mu + 9 = 4\mu + 2 $$

Solve for (\mu)

Rearranging gives: $$ 9 - 2 = 4\mu - 2\mu $$ $$ 7 = 2\mu $$ Thus, $$ \mu = \frac{7}{2} = 3.5 $$

The mean of (X) is $\mu = 3.5$.

More Information

The mean of a random variable is a measure of the center of its distribution. In this case, manipulating the given expected values helped derive the mean effectively. This problem illustrates the use of properties of expected values and equations in statistics.

Tips

Confusing the expected value notation with regular variable calculations.
Not expanding the equations properly, which can lead to incorrect expressions.
Forgetting to account for the constants when manipulating the equations.

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