Question

Let \(X\) be a random variable. If \(m\) is a positive integer, the expectation \(E\left[(X-b)^{m}\right]\), if it exists, is called the \(m\) th moment of the distribution about the point \(b\). Let the first, second, and third moments of the distribution about the point 7 be 3,11, and 15 , respectively. Determine the mean \(\mu\) of \(X\), and then find the first, second, and third moments of the distribution about the point \(\mu\).

Short answer

The mean of \(X\) is 10. The first, second, and third moments of the distribution about the point \(\mu = 10\) are 0, 2, and -19 respectively.

Step-by-step solution

Compute the Mean

To find the mean \(\mu\), use the formula for the first moment about \(b\) which is defined as \(E[X - b] = \mu - b\). Since we know the first moment \(E[X - 7]\) is 3, calculate: \(\mu = 3 + 7 = 10\)

Find the first moment about the Mean

The first moment about the mean \(\mu\) is always zero by definition, since it is \(E[X - \mu]\), which implies: \(E[X - 10] = 0\)

Calculate the second moment about the Mean

Use the formula for the second moment about a point different from the mean, that is \(E[(X - b)^2] = E[(X - \mu)^2] + (\mu - b)^2\). In this case, we have b as 7, \(\mu\) as 10, and \(E[(X - 7)^2]\) as 11. Hence \(E[(X - 10)^2] = 11 - (10 - 7)^2 = 11 - 9 = 2\)

Calculate the third moment about the Mean

Similarly, apply the formula for the third moment about a point different from the mean, that is \(E[(X - b)^3] = E[(X - \mu)^3] + 3(\mu - b)E[(X - \mu)^2] + (\mu - b)^3\). Here \(E[(X - 7)^3]\) is 15, \(\mu\) is 10, b is 7, and \(E[(X - 10)^2]\) found in step 3 is 2. Hence \(E[(X - 10)^3] = 15 - 3(10 - 7)(2) - (10 - 7)^3 = -19\)