Suppose that X_1=delta+W_1 and X_2=2*delta+W_2, where delta, W_1, W_2 are independent standard normal random variables. If the values that we observe happen to be X_1=-1 and X_2=1, then the MAP estimate of delta is?

Answer 1

To find the MAP (Maximum A Posteriori) estimate of delta, we need to consider the conditional probability distribution of delta given the observed values X1 and X2. Let's assume that delta follows a standard normal distribution.

The probability density function (PDF) of X1 can be written as:

f(X1) = (1/sqrt(2*pi))*exp(-0.5*X1^2)

Similarly, the PDF of X2 can be written as:

f(X2) = (1/sqrt(2*pi))*exp(-0.5*X2^2)

Now, we can write the joint probability density function (joint PDF) of X1 and X2 as the product of their respective PDFs, since they are assumed to be independent:

f(X1, X2) = f(X1)*f(X2)

Substituting the values X1 = -1 and X2 = 1, we get:

f(-1, 1) = f(-1)*f(1)

Now, let's express X1 and X2 in terms of delta and W1, W2:

X1 = delta + W1
X2 = 2*delta + W2

Substituting these expressions, we get:

f(-1, 1) = f(delta + W1)*f(2*delta + W2)

Since delta, W1, and W2 are independent standard normal random variables, their PDFs are given by the standard normal distribution:

f(delta) = (1/sqrt(2*pi))*exp(-0.5*delta^2)
f(W1) = (1/sqrt(2*pi))*exp(-0.5*W1^2)
f(W2) = (1/sqrt(2*pi))*exp(-0.5*W2^2)

Now, we can write the joint PDF of delta, W1, and W2 as the product of their respective PDFs:

f(delta, W1, W2) = f(delta)*f(W1)*f(W2)

Substituting these expressions, we get:

f(-1, 1) = (1/sqrt(2*pi))*exp(-0.5*delta^2)*(1/sqrt(2*pi))*exp(-0.5*W1^2)*(1/sqrt(2*pi))*exp(-0.5*W2^2)

Simplifying, we get:

f(-1, 1) = (1/(2*pi))*exp(-0.5*delta^2)*exp(-0.5*W1^2)*exp(-0.5*W2^2)

Now, to find the MAP estimate of delta, we need to find the value of delta that maximizes this joint PDF. Since the term (1/(2*pi)) is a constant, we can ignore it for the maximization:

f(-1, 1) ∝ exp(-0.5*delta^2)*exp(-0.5*W1^2)*exp(-0.5*W2^2)

Taking the logarithm of both sides:

ln(f(-1, 1)) ∝ -0.5*delta^2 - 0.5*W1^2 - 0.5*W2^2

Since we are interested in maximizing this expression, we can ignore any terms that do not depend on delta:

ln(f(-1, 1)) ∝ -0.5*delta^2

Therefore, to find the MAP estimate of delta, we need to find the value of delta that minimizes the expression -0.5*delta^2. This corresponds to maximizing the exponent term, which happens when delta is 0.

Therefore, the MAP estimate of delta, given the observed values X1 = -1 and X2 = 1, is delta = 0.