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butalik [34]
1 year 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]1 year 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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3) You find a jar of quarters on the sidewalk and decide to start collecting them to cash in at the end of the school year.
soldier1979 [14.2K]

Answer:

I dont rally know

Step-by-step explanation:

try it yourself

5 0
1 year ago
Ashley recently opened a store that uses only natural ingredients. She wants to advertise her products by distributing bags of s
Elenna [48]

I will rewrite the question for better understanding:

Ashley recently opened a store that uses only natural ingredients. She wants to advertise her products by distributing bags of samples in her neighborhood. It takes Ashley 2/3 of a minute to prepare one bag. It takes each of her friends 75% longer to prepare a bag. How many hours will it take Ashley and 4 of her friends to prepare 1575 bags of samples?

Answer:

  • <em><u>5.3 hours</u></em>

Explanation:

<u>1) Time it takes Ashley to preprate one bag: </u>

  • 2/3 min

<u>2) Time it takes each friend of Ashley: 75% more than 2/3 min</u>

  • 75% × 2/3 min = 0.75 × 2/3 min = 3/4 × 2/3 min= 2/4 min = 1/2 min = 0.5 min
  • 2/3 min + 1/2 min = 7/6 min

<u>3) Time it takes Ashley and the 4 friends working along to prepare one bag:</u>

  • Convert each time into a rate, since you can set that the total rate of Ashley along with her four friends is equal to the sum of each rate:

  • Rate of Ashley: 1 bag / (2/3) min = 3/2 bag/min

  • Rate of each friend: 1 bag / (7/6) min = 6/7 bag/min

  • Rate of Ashley and 4 friends = 3/2 bag/min + 4 × 6/7 bag/min = (3/2 +24/7) bag/min = 69/14 bag/min

<u>4) Time of prepare 1575 bags of samples:</u>

  • time = number of bags / number of bags per min = 1,575 bags / (69/14) bags/min = 319.56 min

<u>5) Convert minutes to hours:</u>

  • 356.56 min × 1 hour / 60 min = 5.3 hours
3 0
1 year ago
What is 52,634,275,309 in expanded form
DanielleElmas [232]
Fifty-two billion, six hundred and thirty-four million, two hundred and seventy-five thousand, three hundred and nine.
6 0
1 year ago
Read 2 more answers
One similar figure has an area that is nine times the area of another. The larger figure must have dimensions that are times the
frutty [35]
T<span>he area of a figure is the measure of the extent of a two dimensional figure which is usually measured as the square units of the dimensions of the figure.

For example, given a figure with dimensions, say k, the area of the figure is given by k^2.
</span>
<span>Given that o<span>ne similar figure has an area that is nine times the area of another.

Since the two figures are similar, it means that there areas will be proportional as their dimensions will be proportional.

Let the dimensions of the smaller figure be k and the larger figure is p times the smaller figure. Then the area of the smaller figure is k^2 and the area of the larger figure is (pk)^2.

Now, given that the area of the larger figure is nine times the area of the smaller figure, this means that:
\frac{(pk)^2}{k^2} = \frac{9}{1}  \\  \\  \frac{p^2k^2}{k^2} =9 \\  \\ p^2=9 \\  \\ p= \sqrt{9}  \\ \\ p=3
</span>
Therefore, the larger figure must have dimensions that are 3 times the dimensions of the smaller figure.</span>
3 0
1 year ago
Read 2 more answers
You want to make five-letter codes that use the letters A, F, E, R, and M without repeating any letter. What is the probability
Feliz [49]

Answer:

The probability that a randomly chosen code starts with M and ends with E is 0.05 ....

Step-by-step explanation:

According to the given statement we have to make five letter code from  A, F, E, R, and M without repeating any letter. We have to find that what is probability that a randomly chosen code starts with M and ends with E.

Thus the probability of picking the first letter M = 1/5

After that we require the sequence (not E, not E, not E) which is equal to:

= 3/4 * 2/3* *1/2

= 1/4

Now multiply 1/5 and 1/4

1/5 * 1/4

= 1/20

= 0.05

Therefore the probability that a randomly chosen code starts with M and ends with E is 0.05 ....

4 0
2 years ago
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