1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ArbitrLikvidat [17]
3 years ago
15

I need help plz thxzzz

Mathematics
2 answers:
uysha [10]3 years ago
8 0
I will help u but i cant really see what the pic is so can u copy and paste the question on here.
Nonamiya [84]3 years ago
4 0
Hmm... I am not 100% sure but I think the original price was $375
You might be interested in
Let $$X_1, X_2, ...X_n$$ be uniformly distributed on the interval 0 to a. Recall that the maximum likelihood estimator of a is $
Solnce55 [7]

Answer:

a) \hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

b) E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

c) P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

e) On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

Step-by-step explanation:

Part a

For this case we are assuming X_1, X_2 , ..., X_n \sim U(0,a)

And we are are ssuming the following estimator:

\hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

Part b

For this case we assume that the estimator is given by:

E(\hat a) = \frac{na}{n+1}

And using the definition of bias we have this:

E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

And when we take the limit when n tend to infinity we got that the bias tend to 0.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

Part c

For this case we the followng random variable Y = max (X_i) and we can find the cumulative distribution function like this:

P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

Since all the random variables have the same distribution.  

Now we can find the density function derivating the distribution function like this:

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

Now we can find the expected value for the random variable Y and we got this:

E(Y) = \int_{0}^a \frac{n}{a^n} y^n dy = \frac{n}{a^n} \frac{a^{n+1}}{n+1}= \frac{an}{n+1}

And the bias is given by:

E(Y)-a=\frac{an}{n+1} -a=\frac{an-an-a}{n+1}= -\frac{a}{n+1}

And again since the bias is not 0 we have a biased estimator.

Part e

For this case we have two estimators with the following variances:

V(\hat a_1) = \frac{a^2}{3n}

V(\hat a_2) = \frac{a^2}{n(n+2)}

On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

8 0
4 years ago
Witch phrase best defines a a square?
LenaWriter [7]
B. A square has four equal, or congruent, sides. 
3 0
3 years ago
Read 2 more answers
The sum of two consecutive even integers is 122. What are the two numbers?
irakobra [83]

Answer:

  1. Let the first number = x
  2. Therefore, the other number = x + 2 (x+2 is so as in two consecutive even numbers there is 1 odd number)
  3. Now, According to question
  4. x + x + 2 = 122
  5. 2x + 2 = 122
  6. 2x = 122-2
  7. 2x = 120
  8. x = 120/2
  9. x = 60
  10. Now, first no. = x = 60
  11. Therefore, second no. = x + 2 = 60 + 2 = 62
  12. Equation 60 + 62 = 122

Step-by-step explanation:

If you like my answer than please mark me brainliest

7 0
2 years ago
Read 2 more answers
The Idol's Eye diamond weighs 79.2 carats. The Blue Hope diamond weighs 33.68 carats less than the Idol's Eye. What is the weigh
Pavlova-9 [17]
Idol - 33.68 = blue hope
79.2 - 33.68= 45.52 carats <----- Answer
6 0
3 years ago
Read 2 more answers
Choose the statement that is not always true. for a rhombus
andreev551 [17]

Answer:

equal sides


Step-by-step explanation:


the side aren't always the same



6 0
3 years ago
Other questions:
  • Pleassseee helppppppppp
    12·2 answers
  • If tan y/2 = sin y/2 then cos y =?
    10·1 answer
  • Samuel orders four DVDs from an online music store. Each DVD costs $9.99. He has a 20% discount code, and sales tax is 6.75%. Wh
    9·1 answer
  • 7. Two hats and 3 ties cost $88.50. One hat and 2 ties cost $51.00. What does<br> each item cost?
    12·1 answer
  • Help me with this plz
    11·1 answer
  • A triangle has side lengths of 13 MM 18 MM and 14 MM classify it as acute obtuse or right
    8·1 answer
  • 1200 is what percent of 550
    9·2 answers
  • <img src="https://tex.z-dn.net/?f=6%20%7Bx%7D%5E%7B2%7D%20%20-%205x%20-%206" id="TexFormula1" title="6 {x}^{2} - 5x - 6" alt="6
    15·1 answer
  • These cards are put into a bag 1 to 10 one card is chosen at random , what is the probability that the card will have 1 digit on
    14·1 answer
  • how would the fraction 5÷1-√(3 be rewritten if it's denominator is rationalized using difference of squares
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!