Let T1,T2,…,Tn be i.i.d. observations, each drawn from a common normal distribution with mean zero. With probability 1/2 this normal distribution has variance 1, and with probability 1/2 it has variance 4. Based on the observed values t1,t2,…,tn, we use the MAP rule to decide whether the normal distribution from which they were drawn has variance 1 or variance 4. The MAP rule decides that the underlying normal distribution has variance 1 if and only if
∣∣∣c1∑i=1nt2i+c2∑i=1nti∣∣∣<1.
Find the values of c1≥0 and c2≥0 such that this is true. Express your answer in terms of n, and use 'ln' to denote the natural logarithm function, as in 'ln(3)'.
c1=- unanswered
c2=0 - unanswered
0
C1: 3/(8*n*ln(2))
C2: n*ln(1)
c2 = n*ln(1)
please! if you are reading this, be generous and provide your answers for the whole problem set
@RVE, do you know what c1 is equal to? I'm still having trouble with this problem.
anyone know what is c1?
Any clue how you got the term ti, I'm getting only ti^2
Since we are trying to determine variance should it be assumed to be discrete measurement based on continuous observation?
To find the values of c1 and c2 such that the MAP rule decides that the underlying normal distribution has variance 1, we need to evaluate the expression |c1 * Σ(t^2) + c2 * Σ(t)| and set it less than 1.
Let's break down the steps to solve this problem:
Step 1: Compute the sum of squared observations, Σ(t^2), and the sum of observations, Σ(t).
Step 2: Determine the values of c1 and c2 that satisfy the inequality |c1 * Σ(t^2) + c2 * Σ(t)| < 1.
Step 3: Express the values of c1 and c2 using n, the number of observations.
Now let's work through each step:
Step 1: Compute the sum of squared observations, Σ(t^2), and the sum of observations, Σ(t).
- Σ(t^2) is the sum of the squares of the observed values, t1, t2, ..., tn.
- Σ(t) is the sum of the observed values, t1, t2, ..., tn.
Step 2: Determine the values of c1 and c2 that satisfy the inequality |c1 * Σ(t^2) + c2 * Σ(t)| < 1.
- To satisfy the inequality, we need to consider two cases based on the probability distribution:
- Case 1: When the normal distribution has variance 1 (with probability 1/2), the inequality becomes:
|c1 * Σ(t^2) + c2 * Σ(t)| < 1
- Case 2: When the normal distribution has variance 4 (with probability 1/2), the inequality becomes:
|-c1 * Σ(t^2) - c2 * Σ(t)| < 1
- To simplify the notation, we can assume both c1 and c2 are non-negative (c1 ≥ 0 and c2 ≥ 0) since the absolute values account for the case when they are negative.
- We want to find the values of c1 and c2 such that both cases satisfy the respective inequality.
Step 3: Express the values of c1 and c2 using n, the number of observations.
- Since we are asked to express the values of c1 and c2 in terms of n, it means the solution should depend on the number of observations.
At this point, the problem hasn't provided enough information to determine the specific values of c1 and c2 in terms of n. Some additional constraints or information is needed to solve for them.