Consider a fire alarm that senses the environment constantly to figure out if there is smoke in the air and hence to conclude whether there is a fire or not. Consider a simple model for this phenomenon. Let Θ be the unknown true state of the environment: Θ=1 means that there is a fire and Θ=0 means that there is no fire. The signal observed by the alarm at time n is Xn=Θ+Wn, where the random variable Wn represents noise. Assume that Wn is Gaussian with mean 0 and variance 1 and is independent of Θ. Furthermore, assume that for i≠j, Wi and Wj are independent. Suppose that Θ is 1 with probability 0.1 and 0 with probability 0.9.

Give numerical answers for all parts below.

Given the observation X1=0.5, calculate the posterior distribution of Θ. That is, find the conditional distribution of Θ given X1=0.5.

P(Θ=0∣X1=0.5)=

- unanswered



P(Θ=1∣X1=0.5)=

- unanswered

What is the LMS estimate of Θ given X1=0.5?

θˆLMS=

- unanswered

What is the resulting conditional mean squared error of the LMS estimator given X1=0.5?

To calculate the posterior distribution of Θ given X1=0.5, we can use Bayes' theorem. The posterior distribution is given by:

P(Θ=0|X1=0.5) = (P(X1=0.5|Θ=0) * P(Θ=0)) / P(X1=0.5)

P(Θ=1|X1=0.5) = (P(X1=0.5|Θ=1) * P(Θ=1)) / P(X1=0.5)

First, let's calculate the probability of observing X1=0.5 given Θ=0 and Θ=1.

P(X1=0.5|Θ=0) = P(W1=0.5) = 1/sqrt(2π) * e^(-0.5*(0.5-0)^2/1) = 0.352

P(X1=0.5|Θ=1) = P(Θ+W1=0.5|Θ=1) = P(W1=0.5|Θ=1) = 1/sqrt(2π) * e^(-0.5*(0.5-1)^2/1) = 0.127

Next, let's calculate the prior probabilities P(Θ=0) and P(Θ=1). We are given that Θ=1 with probability 0.1 and Θ=0 with probability 0.9.

P(Θ=0) = 0.9
P(Θ=1) = 0.1

Lastly, let's calculate the marginal probability P(X1=0.5) which is the sum of the conditional probabilities weighted by their respective prior probabilities.

P(X1=0.5) = P(X1=0.5|Θ=0) * P(Θ=0) + P(X1=0.5|Θ=1) * P(Θ=1)
= 0.352 * 0.9 + 0.127 * 0.1
= 0.327

Now we can substitute the values into the Bayes' theorem formula to find the posterior distribution.

P(Θ=0|X1=0.5) = (0.352 * 0.9) / 0.327 ≈ 0.968

P(Θ=1|X1=0.5) = (0.127 * 0.1) / 0.327 ≈ 0.032

So the numerical answers are:
P(Θ=0|X1=0.5) ≈ 0.968
P(Θ=1|X1=0.5) ≈ 0.032

The LMS estimate of Θ given X1=0.5 is the value that maximizes the posterior distribution. Since P(Θ=0|X1=0.5) > P(Θ=1|X1=0.5), the LMS estimate is θˆLMS = 0.

The resulting conditional mean squared error of the LMS estimator given X1=0.5 can be calculated using the formula:

MSE = E[(Θ - θˆLMS)^2|X1=0.5]

Since θˆLMS = 0, the MSE reduces to:

MSE = E[Θ^2|X1=0.5]

To find this expected value, we can use the posterior distribution:

MSE = (0^2 * P(Θ=0|X1=0.5)) + (1^2 * P(Θ=1|X1=0.5))

MSE = 0^2 * 0.968 + 1^2 * 0.032

MSE = 0 + 0.032

MSE ≈ 0.032

So the resulting conditional mean squared error of the LMS estimator given X1=0.5 is approximately 0.032.