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butalik [34]
2 years ago
10

We have k coins. The probability of Heads is the same for each coin and is the realized value q of a random variable Q that is u

niformly distributed on [0,1]. We assume that conditioned on Q=q, all coin tosses are independent. Let Ti be the number of tosses of the ith coin until that coin results in Heads for the first time, for i=1,2,…,k. (Ti includes the toss that results in the first Heads.)
You may find the following integral useful: For any non-negative integers k and m,

∫10qk(1−q)mdq=k!m!(k+m+1)!.
Find the PMF of T1. (Express your answer in terms of t using standard notation.)

For t=1,…, pT1(t)=- unanswered
Find the least mean squares (LMS) estimate of Q based on the observed value, t, of T1. (Express your answer in terms of t using standard notation.)

E[Q∣T1=t]=- unanswered
We flip each of the k coins until they result in Heads for the first time. Compute the maximum a posteriori (MAP) estimate q^ of Q given the number of tosses needed, T1=t1,…,Tk=tk, for each coin.
Mathematics
1 answer:
nata0808 [166]2 years ago
3 0

Answer:

Step-by-step explanation:

The question is not in correct order ; so the first thing we are required to do is to put them together in the right form to make it easier to proof; having said that. let's get started!.

From the second part " You may find the following integral useful: For any non-negative integers k and m,"

The next  equation goes thus : \int\limits^1_0 \ \  q^k (1-q)^m dq = \frac{k!m!}{(k+m+1)!}

a. Find the PMF of T_1 .   (Express your answer in terms of t using standard notation.)

For t=1,2…,

p_T_1(t)= \frac{1}{(t*(t*1))}  by using conditional probability to solve PMF

b)  Find the least mean squares (LMS) estimate of Q based on the observed value, t, of T1. (Express your answer in terms of t using standard notation.)

By using the mathematical definition of integration to solve the estimate of Q:

E|Q|T_1=t|= \frac{2}{t+2}

c)  We flip each of the k coins until they result in Heads for the first time. Compute the maximum a posteriori (MAP) estimate q^ of Q given the number of tosses needed, T1=t1,…,Tk=tk, for each coin.

The MAP estimate of  q is :

\bar q = \frac{k}{\sum^k_{i=1}}t_i

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Quadrilaterals WXYZ and BADC are congruent. In addition, WX ≅ DC and XY ≅ BC. If AD = 4 cm and AB = 6 cm, what is the perimeter
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If <u>quadrilaterals</u> WXYZ and BADC are <u>congruent</u>, then their corresponding <u>sides</u> are congruent.

Given that

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you can state that

  • YZ≅AB,
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If AD = 4 cm and AB = 6 cm, then WZ = 4 cm and YZ = 6 cm. Opposite rectangle sides are congruent, then XY = 4 cm and WX = 6 cm.

The perimeter of WXYZ is

P = WX + XY + YZ + WZ = 6 + 4 + 6 + 4 = 20 cm.

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Let s be the distance traveled the second day.

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Answer:

The coordinates for the location of the center of the platform are (0, 1)

Step-by-step explanation:

The equation of the circle of center (h , k) and radius r is:

(x - h)² + (y - k)² = r²

Now,

- The center is equidistant from any point lies on the circumference of the circle

- There are three points equidistant from the center of the circle

- We have three unknowns in the equation of the circle h , k , r

Thus, let's substitute the coordinates of these point in the equation of the circle to find h , k , r.

The equation of the circle is (x - h)² + (y - k)² = r²

∵ Points A(2,−3), B(4,3), and C(−2,5)

- Substitute the values of x and y the coordinates of these points

Point A (2 , -3)

(2 - h)² + (-3 - k)² = r² - - - (1)

Point B (4 , 3)

(4 - h)² + (3 - k)² = r² - - - - (2)

Point C (-2 , 5)

(-2 - h)² + (5 - k)² = r² - - - - (3)

- To find h , k equate equation (1) and (2) and same for equation (2) and (3) because all of them equal r²

Thus;

(2 - h)² + (-3 - k)² = (4 - h)² + (3 - k)² - - - - - (4)

(4 - h)² + (3 - k)² = (-2 - h)² + (5 - k)² - - - - -(5)

- Simplify (5);

h² - 8h + 16 + k² - 6k + 9 = h² + 4h + 4 + k² - 10k + 25

h² and k² will cancel out to give;

-8h - 6k + 25 = 4h - 10k + 29

Rearranging, we have;

12h - 4k = -4 - - - - (6)

Similarly, for equation 4;

(2 - h)² + (-3 - k)² = (4 - h)² + (3 - k)²

h² - 4h + 4 + k² + 6k + 9 = h² - 8h + 16 + k² - 6k + 9

h², k² and 9 will cancel out to give;

4 - 4h + 6k = 16 - 8h - 6k

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4h + 12k = 12 - - - - (7)

Divide by 4 to give;

h + 3k = 3

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h = 3 - 3k

Put 3 - 3k for h in eq 6;

12(3 - 3k) - 4k = -4

36 - 36k - 4k = -4

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k = 40/40

k = 1

h = 3 - 3(1)

h = 0

The coordinates for the location of the center of the platform are (0, 1)

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