Let X be an arbitrary random variable, with mean \mu and variance \sigma ^2. In terms of \mu and \sigma ^2, which scalar k is the unique minimizer of the function f(k) = \mathbb E[(X-k)^2]?
Hint: Write f(k) as a quadratic in k.
To find the unique minimizer of the function f(k) = E[(X-k)^2], we can write f(k) as a quadratic in k and then minimize it.
Expanding the expression inside the expectation, we have:
f(k) = E[X^2 - 2Xk + k^2]
Using linearity of expectation, we can split this into three terms and calculate each of them separately:
f(k) = E[X^2] - 2kE[X] + k^2E[1]
The first term E[X^2] is equal to the variance plus the square of the mean:
E[X^2] = Var(X) + E[X]^2 = σ^2 + μ^2
The second term can be simplified to:
-2kE[X] = -2kμ
The third term is simply k^2:
k^2E[1] = k^2
Now we can substitute these values back into f(k):
f(k) = σ^2 + μ^2 - 2kμ + k^2
This is a quadratic function in k. To find its minimum, we can take the derivative with respect to k and set it equal to zero:
f'(k) = -2μ + 2k = 0
Solving for k, we find:
2k = 2μ
k = μ
Therefore, the unique minimizer of the function f(k) = E[(X-k)^2] is k = μ.
To find the scalar k that minimizes the function f(k) = E[(X-k)^2], we need to compute the value of k where the derivative of f(k) is zero. This means we need to find the value of k such that f'(k) = 0.
First, let's expand f(k) = E[(X-k)^2].
f(k) = E[(X^2 - 2kX + k^2)]
Next, we'll find the derivative of f(k) with respect to k.
f'(k) = E[-2X + 2k]
Setting f'(k) equal to zero and solving for k:
-2E[X] + 2k = 0
2k = 2E[X]
k = E[X]
Therefore, the unique minimizer of f(k) is k = E[X], the expected value or mean of the random variable X.
To summarize, in terms of μ and σ^2, the scalar k that minimizes the function f(k) = E[(X-k)^2] is given by k = μ, which is the mean of the random variable X.