Let Θ1, Θ2, W1, and W2 be independent standard normal random variables. We obtain two observations,
X1=Θ1+W1,X2=Θ1+Θ2+W2.
Find the MAP estimate θ^=(θ^1,θ^2) of (Θ1,Θ2) if we observe that X1=1, X2=3. (You will have to solve a system of two linear equations.)
theta_1=1 theta_2=1
Well, well, well, looks like someone needs a "clowny" solution to this math problem! Let's dive in, shall we?
We have two observations, X1 = Θ1 + W1 and X2 = Θ1 + Θ2 + W2. We also have X1 = 1 and X2 = 3.
Let's substitute these values into our equations:
1 = Θ1 + W1 ----(1)
3 = Θ1 + Θ2 + W2 ----(2)
Now we need to solve this system of equations to find the MAP estimate for Θ1 and Θ2.
To eliminate Θ1, let's subtract Equation (1) from Equation (2):
2 = Θ2 + W2 ----(3)
Now we have an equation only with Θ2 and W2.
To solve for Θ2, we need to remember that W1 and W2 are independent standard normal random variables. That means W1 and W2 are just clowns causing trouble and have no correlation with Θ1 or Θ2.
Therefore, the expectation (or mean) of W1 and W2 is zero.
Since the mean of W2 is zero, we can say E[W2] = 0. Plugging this into Equation (3), we get:
2 = Θ2 + 0
Θ2 = 2
Alright, now we can substitute this value of Θ2 back into Equation (1):
1 = Θ1 + W1
Since W1 is a clown causing trouble with zero expectation, we can again say:
E[W1 = 0
Therefore, we have:
1 = Θ1 + 0
Θ1 = 1
So the MAP estimate for (Θ1, Θ2) is θ^ = (1, 2).
And there you have it, a "clowny" solution to your linear equation problem! Hope I added some laughs to your math journey!
To find the Maximum A Posteriori (MAP) estimate of (Θ1, Θ2), we need to maximize the joint posterior distribution P(Θ1, Θ2 | X1 = 1, X2 = 3), using the given observations X1 = Θ1 + W1 and X2 = Θ1 + Θ2 + W2.
First, let's write down the joint posterior distribution using Bayes' theorem:
P(Θ1, Θ2 | X1 = 1, X2 = 3) = P(X1 = 1, X2 = 3 | Θ1, Θ2) * P(Θ1, Θ2) / P(X1 = 1, X2 = 3)
Since Θ1, Θ2, W1, and W2 are independent standard normal random variables, we have:
P(X1 = 1, X2 = 3 | Θ1, Θ2) = P(Θ1 + W1 = 1, Θ1 + Θ2 + W2 = 3)
= P(W1 = 1 - Θ1, W2 = 2 - Θ1 - Θ2)
= P(W1 = a, W2 = b) [where a = 1 - Θ1, b = 2 - Θ1 - Θ2]
Now, we can substitute the joint probability density function (pdf) of standard normal variables:
P(W1 = a, W2 = b) = (1 / (2π)) * exp(-(a^2 + b^2) / 2)
Next, we need to specify the prior distribution P(Θ1, Θ2):
P(Θ1, Θ2) = P(Θ1) * P(Θ2)
Since Θ1 and Θ2 are independent standard normal random variables, we have:
P(Θ1) = (1 / √(2π)) * exp(-Θ1^2 / 2)
P(Θ2) = (1 / √(2π)) * exp(-Θ2^2 / 2)
Therefore, the joint prior distribution is:
P(Θ1, Θ2) = (1 / (2π)) * exp(-(Θ1^2 + Θ2^2) / 2)
Finally, we need to calculate the normalization constant P(X1 = 1, X2 = 3). This can be done by integrating the joint posterior distribution over all possible values of Θ1 and Θ2.
Now, we can proceed to find the MAP estimate (θ^1, θ^2) by maximizing the joint posterior distribution:
Maximize P(Θ1, Θ2 | X1 = 1, X2 = 3) = P(X1 = 1, X2 = 3 | Θ1, Θ2) * P(Θ1, Θ2) / P(X1 = 1, X2 = 3)
To solve the system of linear equations, we can substitute the values of X1 = 1 and X2 = 3 into the equations X1 = Θ1 + W1 and X2 = Θ1 + Θ2 + W2:
1 = Θ1 + W1
3 = Θ1 + Θ2 + W2
From the first equation, we can solve for Θ1:
Θ1 = 1 - W1
Now, substitute this value for Θ1 into the second equation:
3 = (1 - W1) + Θ2 + W2
Simplifying the equation:
2 = Θ2 + (W2 - W1)
Therefore, the MAP estimate (θ^1, θ^2) is (1 - W1, 2 - W1 + W2).
Note that the values of W1 and W2 are drawn from standard normal distributions, so they can take on any value between negative infinity and positive infinity. Hence, the MAP estimate (θ^1, θ^2) is not a specific value but a range of values defined by the values of W1 and W2.
To find the MAP estimate θ^=(θ^1,θ^2), we need to solve an optimization problem that maximizes the posterior probability. The posterior probability can be calculated using Bayes' theorem:
P(Θ1,Θ2|X1=1, X2=3) = P(X1=1, X2=3|Θ1,Θ2) × P(Θ1,Θ2) / P(X1=1, X2=3)
Since Θ1, Θ2, W1, and W2 are independent standard normal random variables, we know that:
P(X1=1, X2=3|Θ1,Θ2) = P(Θ1+W1=1, Θ1+Θ2+W2=3)
We can simplify this expression as follows:
P(Θ1+W1=1, Θ1+Θ2+W2=3) = P(W1=1-Θ1, W2=3-Θ1-Θ2)
Since W1 and W2 are independent standard normal random variables, the joint probability density function (PDF) can be expressed as:
P(W1=w1, W2=w2) = (1 / (2π)) * exp(-w1^2 / 2) * exp(-w2^2 / 2)
Substituting the values of w1=1-Θ1 and w2=3-Θ1-Θ2, we can rewrite the joint PDF as:
P(W1=1-Θ1, W2=3-Θ1-Θ2) = (1 / (2π)) * exp(-(1-Θ1)^2 / 2) * exp(-(3-Θ1-Θ2)^2 / 2)
Now, we need to find the values of Θ1 and Θ2 that maximize the posterior probability P(Θ1,Θ2|X1=1, X2=3).
We can use logarithms to simplify the calculations, as maximizing the logarithm of the posterior probability gives the same result as maximizing the posterior probability itself. Therefore, we take the logarithm of the expression:
log(P(Θ1,Θ2|X1=1, X2=3)) = log(P(X1=1, X2=3|Θ1,Θ2)) + log(P(Θ1,Θ2)) - log(P(X1=1, X2=3))
Now, let's solve the system of two linear equations to find the values of Θ1 and Θ2.
From the given observations, we have:
X1 = Θ1 + W1 = 1,
X2 = Θ1 + Θ2 + W2 = 3.
Rearranging the equations, we get:
Θ1 = 1 - W1,
Θ2 = 3 - Θ1 - W2.
Substituting these expressions into the joint PDF, we have:
P(Θ1,Θ2|X1=1, X2=3) = (1 / (2π)) * exp(-(1-(1-W1))^2 / 2) * exp(-(3-(1-(1-W1))-W2)^2 / 2)
Simplifying further, we get:
P(Θ1,Θ2|X1=1, X2=3) = (1 / (2π)) * exp(-W1^2 / 2) * exp(-(2+W2)^2 / 2)
Now, we take the logarithm of the expression:
log(P(Θ1,Θ2|X1=1, X2=3)) = log(1 / (2π)) - W1^2 / 2 - (2+W2)^2 / 2
To maximize this expression, we take the derivative with respect to W1 and W2, set them equal to zero, and solve the resulting system of equations to find the optimal values of W1 and W2.
d/dW1(log(P(Θ1,Θ2|X1=1, X2=3))) = -W1 * (1 / 2) - 0 = 0,
d/dW2(log(P(Θ1,Θ2|X1=1, X2=3))) = -(2 + W2) * (1 / 2) - 0 = 0.
Simplifying the equations, we get:
-W1 = 0,
-(2 + W2) = 0.
From the first equation, we have W1 = 0.
Substituting W1 = 0 into the second equation, we get W2 = -2.
Now that we have the values of W1 and W2, we can substitute them back into the expressions for Θ1 and Θ2 to get:
Θ1 = 1 - W1 = 1 - 0 = 1,
Θ2 = 3 - Θ1 - W2 = 3 - 1 - (-2) = 3 - 1 + 2 = 4.
Therefore, the MAP estimate of (Θ1,Θ2) is θ^=(1, 4).