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3 years ago

6

hi whoever is reading this, please consider helping me. click on this. I know the answer is 19 because a is 18 and b is 1 but id

k how to solve it algebraically.

1 answer:

babymother [125]3 years ago

5 0

Answer:

A= 108

Step-by-step explanation:

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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

I need a little help with this problem: 5k + 1 = 4k - 1

Varvara68 [4.7K]

5k-4k=-1+-1
k=-2, its all aboit rearranging the order of numders based on there like terms. If you need any more help ask.

8 0

3 years ago

Brian correctly use a method of completing the square to solve the equation X a 2nd plus 7X -11 equals zero Brian‘s first step w

ArbitrLikvidat [17]

Answer:

$(\frac{7}{2})^{2}$

Step-by-step explanation:

Brain correctly use a method of completing the square to solve the equation:

$x^2+7x-11=0$

His First Step is to: Take the Constant Term to the Right Hand Side

$x^2+7x=11$

The Next Step Would be to:

Divide the Coefficient of x by 2
Square It
Add it to both Sides

In this case, the Coefficient of x = 7

Divided by 2 = $\frac{7}{2}$

Squaring It, we have: $(\frac{7}{2})^{2}$

It is this number $(\frac{7}{2})^{2}$ that is added to both sides in the manner below:

$x^2+7x+(\frac{7}{2})^{2}=11+(\frac{7}{2})^{2}$

6 0

4 years ago

Joe is 5 years older than Sydney. The sum of their ages is 45. What is Sydney’s age?

sweet-ann [11.9K]

Answer:

20 years old

Step-by-step explanation:

Joe = J

Sydney = S

J = 5 + S

J + S = 45

Substitute

5 + S + S = 45

Add

5 + 2S = 45

Subtract 5 from both sides of the equation

2S = 40

Divide both sides of the equation by 2

S = 20

Sydney is 20 years old

Hope this helps :)

8 0

3 years ago

Which expression can be expanded using the binomial theorem?

kirill [66]

Answer:

The third choice: (x² - 1)³

Step-by-step explanation:

That is the only expression that is a binomial. Binomial means "two numbers". For this expression, x² is one number, and -1 is the other. Choice 1 is a monomial (one number) and choices 2 and 4 are trinomials (three numbers)

7 0

3 years ago