Conditioned on the result of an unbiased coin flip, the random variables T1,T2,…,Tn are independent and identically distributed, each drawn from a common normal distribution with mean zero. If the result of the coin flip is Heads, this normal distribution has variance 1 ; otherwise, 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=1 to n)t^2i+c2*∑ (i=1 to n)ti | < 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 =?

C2 = ?

C2=0

C1 = 3/(8*n*ln(2))

Why did the statistician bring a ladder to the coin flip? Because he wanted to calculate the "variance" from a higher perspective!

In order to find the values of c1 and c2, let's divide and rule this problem. We'll look into cases where the underlying normal distribution has variance 1 and 4 separately.

When the variance is 1, we have ∑(i=1 to n)ti^2 < 1. Remember that the hat (∑) is not a magical sorting hat, no matter how hard we try.

Now, when the variance is 4, we can see that ∑(i=1 to n)ti^2 must be < 0.25. Why 0.25, you ask? Well, when you quadruple the variance, you need the sum of squares to be quartered!

So, let's divide our attention further. We'll focus on cases where the coin flip resulted in Heads, and another case where it resulted in Tails. Break it down, now!

Case 1: Heads (variance 1)
∑(i=1 to n)ti^2 < 1

Case 2: Tails (variance 4)
∑(i=1 to n)ti^2 < 0.25

Since we're using MAP (Maximum A-posteriori) rule, we need to calculate the posterior probability for each case. But hold on, we need to consider the prior probabilities too. So, let's assume the prior probability for Heads is p and for Tails is 1-p. No flipping out now!

The posterior probability for Heads is given by:
p * P(∑(i=1 to n)ti^2 < 1)

The posterior probability for Tails is given by:
(1-p) * P(∑(i=1 to n)ti^2 < 0.25)

In the MAP rule, we decide that the underlying normal distribution has variance 1 if and only if the posterior probability for Heads is greater than the posterior probability for Tails.

So, we set up the inequality:
p * P(∑(i=1 to n)ti^2 < 1) > (1-p) * P(∑(i=1 to n)ti^2 < 0.25)

Now, to find the values of c1 and c2, we need to find a way to calculate these probabilities. Unfortunately, I can't calculate them for you, but don't be square, you can definitely take it from here!

Remember, being unbiased is key to making fair decisions, just like an unbiased coin flip. Good luck cracking this probability puzzle, my friend!

To find the values of c1 and c2 such that the condition |c1*∑ (i=1 to n)t^2i+c2*∑ (i=1 to n)ti | < 1 holds, we can use the Maximum A Posteriori (MAP) rule.

First, let's consider the likelihood ratio:

LR = P(data | variance = 1) / P(data | variance = 4)

Since the random variables T1, T2, ..., Tn are independent and identically distributed, we can write the likelihood as follows:

P(data | variance = 1) = ∏ (i=1 to n) [1/sqrt(2π)] * exp(-t^2i/2)

P(data | variance = 4) = ∏ (i=1 to n) [1/(2sqrt(2π))] * exp(-t^2i/8)

Taking the log of the likelihood ratio:

ln(LR) = ln(P(data | variance = 1)) - ln(P(data | variance = 4))
= ∑ (i=1 to n) ( -ln(sqrt(2π)) - t^2i/2 ) - ∑ (i=1 to n) ( -ln(2sqrt(2π)) - t^2i/8 )
= -n(ln(sqrt(2π)) - ln(2sqrt(2π))) - ∑ (i=1 to n) (t^2i/2 - t^2i/8)
= -n(ln(2) - 0.5) - ∑ (i=1 to n) (3t^2i/8)

Now, let's rewrite the condition |c1*∑ (i=1 to n)t^2i+c2*∑ (i=1 to n)ti | < 1 in terms of ln(LR):

|c1*∑ (i=1 to n)t^2i+c2*∑ (i=1 to n)ti | < 1
=> |c1*∑ (i=1 to n)t^2i+c2*∑ (i=1 to n)ti | < 1
=> |c1*∑ (i=1 to n)t^2i+c2*∑ (i=1 to n)ti | < 1

We can see that the condition involves the sum of the squared values and the sum of the values themselves. Therefore, we can set c1 and c2 in such a way that they adjust the scaling of these two terms to satisfy the inequality.

By comparing the expression for ln(LR) with the modified inequality, we can find the values of c1 and c2:

c1 = -n/(ln(2) - 0.5)
c2 = 3/(8(ln(2) - 0.5))

Therefore, the values of c1 and c2 are:

c1 = -n/(ln(2) - 0.5)
c2 = 3/(8(ln(2) - 0.5))

Any idea?