A Gaussian density has mean µ and variance σ^2:

p(x|µ, σ) = 1/√(2πσ^2)exp (−(x − µ)^2/2σ^2)
If y = (x − µ)^2
a) What is the density function of y?
b) What value of y has highest density?

To determine the density function of y, we need to find the distribution of the random variable y, defined as y = (x - µ)^2, where x follows a Gaussian density with mean µ and variance σ^2.

a) To find the density function of y, we need to find the cumulative distribution function (CDF) of y and then differentiate it to obtain the probability density function (PDF).

Let's start by finding the CDF of y, denoted as F(y). We want to find the probability that y is less than or equal to a given value y0:

F(y0) = P(y ≤ y0)

From the definition of y, we can rewrite the inequality as:

(x - µ)^2 ≤ y0

Taking the square root of both sides, we get:

|x - µ| ≤ √y0

Now, since x follows a Gaussian density with mean µ and variance σ^2, we can write the above inequality in terms of x:

-√y0 + µ ≤ x ≤ √y0 + µ

Using the cumulative distribution function (CDF) of the Gaussian density, denoted as Φ, we can express the probability as:

F(y0) = P(x ≤ √y0 + µ) - P(x ≤ -√y0 + µ) = Φ(√y0 + µ) - Φ(-√y0 + µ)

Differentiating the CDF with respect to y0 gives us the PDF of y:

f(y) = dF(y)/dy = d(Φ(√y0 + µ))/dy - d(Φ(-√y0 + µ))/dy

Now, we need to use the chain rule to differentiate Φ with respect to y0:

dΦ(√y0 + µ)/dy0 = dΦ(u)/du * du/dy0 = dΦ(u)/du * (1/2√y0)

Similarly, differentiating Φ(-√y0 + µ) with respect to y0, we get:

dΦ(-√y0 + µ)/dy0 = dΦ(v)/dv * (1/-2√y0)

Substituting these expressions into the PDF expression, we have:

f(y) = dΦ(u)/du * (1/2√y0) - dΦ(v)/dv * (1/-2√y0)

Therefore, the density function of y is given by:

f(y) = (1/2√y0) * (dΦ(u)/du - dΦ(v)/dv)

where u = √y0 + µ and v = -√y0 + µ.

b) Now, to find the value of y that has the highest density, we need to find the maximum point of the density function f(y) that we derived in part a).

To find the maximum, we need to set the derivative of f(y) with respect to y equal to zero and solve for y. However, this may not lead to a straightforward solution as it involves differentiating the CDF of the Gaussian density, which does not have a simple closed-form expression.

Therefore, a common approach is to find the mean of the distribution of y, denoted as E[y], rather than finding the specific value of y with the highest density. The mean of y can be obtained by taking the expectation of y over its distribution:

E[y] = ∫ y * f(y) dy

To calculate this integral, we would need to evaluate the derivative of the CDF of the Gaussian density (Φ) and perform the integral, but this can be quite involved.

Alternatively, we can note that since y is the squared difference between x and µ, it cannot be negative. Therefore, y has a lower bound of zero. Moreover, as the variance σ^2 approaches zero, the density of x becomes more concentrated around the mean µ. In this limit, y would also become more concentrated around zero. Hence, we can conclude that the value of y with the highest density is y = 0.