The maximum likelihood estimator θ-hat is biased, but the bias tends to 0 as the sample size increases. It is not a minimum variance unbiased estimator.
Given, Y1, Y2, …, Ym are the random samples of size m from Gamma (α=3, β=θ), where θ is not known and the aim is to check whether or not the maximum likelihood estimator θ-hat is a minimum variance unbiased estimator of the parameter θ.
Now, let's calculate the maximum likelihood estimator of θ. The density function of Gamma distribution is given by: f(y|θ) = (1/Γ(α) βα) yα-1 e-y/β. The likelihood function of θ is given by: L(θ) = (1/Γ(α) βα)m Πi=1myiα-1 e-yi/β. We can take the natural logarithm of this to obtain: l(θ) = m log (1/Γ(α)) + mα log(β) + (α-1) Σi=1mlog (yi) - (1/β) Σi=1myi.
Now, we differentiate this with respect to θ to obtain: d/dθl(θ) = mα/θ - Σi=1myi/θ = 0. Therefore,θ-hat = (1/m) Σi=1myi = Y-bar. The expected value of θ-hat is given by: E(θ-hat) = E(Y-bar) = (1/m) Σi=1m E(Yi) = (1/m) Σi=1m (αβ) = αβ=3θ. Thus, the maximum likelihood estimator θ-hat is biased but the bias tends to 0 as m tends to infinity.
Now, let's calculate the variance of θ-hat. The variance of θ-hat is given by:
Var(θ-hat) = Var(Y-bar) = (1/m²) Σi=1mVar(Yi) = αβ/m = 3θ/m. Thus, the mean square error (MSE) of θ-hat is given by: MSE(θ-hat) = Var(θ-hat) + Bias²(θ-hat) = (3θ/m) + 0
Thus, the maximum likelihood estimator θ-hat is not a minimum variance unbiased estimator of the parameter θ. Therefore, it can be concluded that the estimator is biased.
For more questions on minimum variance
https://brainly.com/question/14893757
#SPJ8
Compute the sample odds if p=0.45. For credit vour answer must be a decimal, accurate to two decimal places. If your answer is less than 1 there must be a zero before the decimal p
In probability theory, odds are the chances of an event taking place. The odds of an event can be calculated by dividing the probability of an event occurring by the probability of it not occurring.
Sample odds refer to the odds derived from the sample data.Compute the sample odds if p = 0.45If p = 0.45, the probability of the event occurring is 0.45. The probability of the event not occurring is 1 - 0.45 = 0.55.The odds of the event occurring are given by:P(event occurring) / P(event not occurring) = 0.45 / 0.55 = 0.8182 (rounded to four decimal places)
To convert this to a decimal accurate to two decimal places, we round the answer to 0.82.So the sample odds are 0.82. Hence, the is 0.82.This is a, but it satisfies the criteria mentioned above as it is accurate to two decimal places and provides the required format.
To know more about probability visit :
https://brainly.com/question/31828911
#SPJ11
In the Introduction to Impossibility module, 4/5 of students pass the exam. Of those who pass the exam 9/10 attend lectures, and of those who fail only 1/2 attend lectures. What is the probability that a student who doesn't attend lectures passes the exam? Select one: a. None of the other choices b. 1/10 c. 4/9 d. 2/25 e. 1/2
The probability that a student who doesn't attend lectures passes the exam is 2/25.
Let's assume that there are 100 students in total. From the given information, we know that 4/5 of the students (80 students) pass the exam, and 1/5 of the students (20 students) fail the exam.
Out of the 80 students who pass the exam, 9/10 of them attend lectures. Therefore, the number of students who pass the exam and attend lectures is (9/10) * 80 = 72.
Out of the 20 students who fail the exam, only 1/2 of them attend lectures. So, the number of students who fail the exam and attend lectures is (1/2) * 20 = 10.
Now, we need to find the probability that a student who doesn't attend lectures passes the exam. This means we need to find the probability of passing the exam among the students who do not attend lectures, which is the complement of attending lectures.
The number of students who do not attend lectures is 100 - 72 - 10 = 18. Out of these 18 students, only 2 students pass the exam.
Therefore, the probability that a student who doesn't attend lectures passes the exam is 2/18, which simplifies to 1/9. In decimal form, this is approximately 0.08, which can be further simplified to 2/25.
Thus, the correct answer is option d) 2/25.
Learn more about probability here:
https://brainly.com/question/31828911
#SPJ11
Solve the system by using Gaussian elimination or Gauss-Jordan elimination.
5x−19y+12z=−110
x−4y+3z=−23
2x−3y−7z=−19
we can see that the system is consistent and has a unique solution. The solution to the system is x = 242, y = -132, and z = -1.
To solve the given system of equations using Gaussian elimination or Gauss-Jordan elimination, we will perform row operations to transform the augmented matrix into row-echelon form or reduced row-echelon form.
The augmented matrix representing the system is:
[5 -19 12 | -110]
[1 -4 3 | -23]
[2 -3 -7 | -19]
First, we will perform row operations to create zeros below the first entry in the first column. We can achieve this by multiplying the second row by 5 and adding it to twice the third row. The resulting matrix is:
[5 -19 12 | -110]
[0 1 -2 | -130]
[0 7 -13 | -267]
Next, we will perform row operations to create zeros below the second entry in the second column. We can achieve this by multiplying the second row by 7 and adding it to seven times the third row. The resulting matrix is:
[5 -19 12 | -110]
[0 1 -2 | -130]
[0 0 1 | -1]
Now, we will perform row operations to create zeros above the third entry in the third column. We can achieve this by adding twice the third row to the second row and adding -12 times the third row to the first row. The resulting matrix is:
[5 -19 0 | 2]
[0 1 0 | -132]
[0 0 1 | -1]
Finally, we will perform row operations to create zeros above the second entry in the second column. We can achieve this by adding 19 times the second row to the first row. The resulting matrix is:
[5 0 0 | 242]
[0 1 0 | -132]
[0 0 1 | -1]
From the row-echelon form of the augmented matrix, we can see that the system is consistent and has a unique solution. The solution to the system is x = 242, y = -132, and z = -1.
Learn more about Gauss-Jordan elimination,here:
brainly.com/question/30767485
#SPJ11
The Earth is approximately a sphere of radius 6.37 x 10% m. Calculate the distance from the pole to the equator, measured along the surface of the Earth. Calculate the distance from the pole to the equator, measured along a straight line passing through the Earth.
The distance from the pole to the equator, measured along the surface of the Earth, is approximately 10,002 km. The distance from the pole to the equator, measured along a straight line passing through the Earth, is equal to the diameter of the Earth, which is approximately 12,740 km.
To calculate the distance from the pole to the equator along the surface of the Earth, we can use the circumference formula for a circle:
C = 2πr
Given that the radius of the Earth is 6.37 x 10^6 m, the distance along the surface of the Earth from the pole to the equator is:
C = 2π(6.37 x 10^6 m) ≈ 40,074 km
However, since we are measuring along the surface of the Earth, we need to consider only half of the circumference, which gives us:
Distance = 1/2 * C ≈ 20,037 km
To calculate the distance from the pole to the equator along a straight line passing through the Earth, we use the diameter of the Earth. The diameter is twice the radius, so:
Distance = 2 * (6.37 x 10^6 m) ≈ 12,740 km
Therefore, the distance from the pole to the equator, measured along the surface of the Earth, is approximately 10,002 km, while the distance measured along a straight line passing through the Earth (the diameter) is approximately 12,740 km.
Learn more about ccircumference hhere:
https://brainly.com/question/28757341
#SPJ11
1.
Sketch the graph of the polynomial function given by: y=x(x-2)x+3)(x-5)
(a) LIST all the zeroes as ordered pairs and, (b) DESCRIBE the end behaviors of the function.
2. Sketch the graph of the rational function given by: y 3-2x x+1 LABEL the vertical asymptote(s) and any the horizontal asymptote with their equations, and the x-and y- intercept(s). SHOW HOW YOU ARRIVE AT YOUR VALUES!!
1.(a) The zeroes of the function given by y = x(x-2)(x+3)(x-5) are of the form (x,0).Therefore, to find all the zeroes of the function, we equate the function to zero and solve for x:y = x(x-2)(x+3)(x-5) = 0∴ x = 0, 2, -3, 5The ordered pairs are (0,0), (2,0), (-3,0) and (5,0).(b) The leading term in the polynomial function y = x(x-2)(x+3)(x-5) is x⁴.
This has a positive coefficient, therefore the end behavior of the function is that of a fourth-degree polynomial function with a positive leading coefficient, i.e. as x → ∞ and x → -∞, the value of the function y increases without bound.2. Given the rational function y = (3 - 2x)/(x+1)
To find the vertical asymptotes of the function, we look for all the values of x for which the denominator is zero, since division by zero is undefined. In this case, the denominator is (x+1) and so the vertical asymptote is x = -1.
Therefore, the equation of the vertical asymptote is x = -1.To find the horizontal asymptote of the function, we use the fact that the degree of the numerator (which is 1) is less than the degree of the denominator (which is 2).
Therefore, the horizontal asymptote is y = 0.To find the x-intercept of the function, we set y = 0 and solve for x:0 = (3 - 2x)/(x+1)⇒ 0 = 3 - 2x⇒ 2x = 3⇒ x = 3/2 Therefore, the x-intercept is (3/2, 0).To find the y-intercept of the function, we set x = 0 and solve for y:y = (3 - 2x)/(x+1)⇒ y = 3
Learn more about polynomial
https://brainly.com/question/11536910
#SPJ11
A moving particle starts at an initial position r(0) = (0, 1, 1) with initial velocity v(0) = (0, 0, 1). Its acceleration at any given time t is a(t) = (t, e^t, e^-t). Find its velocity and position vectors at any given time t.
Given that a particle starts at an initial position r(0) = (0,1,1) with initial velocity v(0) = (0,0,1) and its acceleration at any given time t is a(t) = (t, e^t, e^-t). We need to find the velocity and position vectors at any given time t.
Velocity Vector:The acceleration vector is given as a(t) = (t, e^t, e^-t)We know that acceleration is the derivative of velocity with respect to time.So, integrating a(t) w.r.t time we get velocity vector v(t) as: v(t) = (t²/2, e^t, -e^-t) + c1Where c1 is the constant of integration.Using the initial condition given, i.e., v(0) = (0,0,1) we can calculate the value of c1.Substituting v(0) = (0,0,1), we get c1 = (0,0,0)Hence, the velocity vector at any given time t is given as v(t) = (t²/2, e^t, -e^-t)Conclusion: Therefore, the velocity vector at any given time t is v(t) = (t²/2, e^t, -e^-t)Position Vector:The velocity vector is given as v(t) = (t²/2, e^t, -e^-t)We know that velocity is the derivative of position vector with respect to time.So, integrating v(t) w.r.t time we get the position vector r(t) as: r(t) = (t³/6, e^t, e^-t) + c2Where c2 is the constant of integration.
Using the initial condition given, i.e., r(0) = (0,1,1) we can calculate the value of c2.Substituting r(0) = (0,1,1), we get c2 = (0,1,1)Hence, the position vector at any given time t is given as r(t) = (t³/6, e^t+1, e^-t+1)
Therefore, the velocity vector at any given time t is v(t) = (t²/2, e^t, -e^-t) and the position vector at any given time t is r(t) = (t³/6, e^t+1, e^-t+1).
To know more about initial Visit
https://brainly.com/question/32209767
#SPJ11
numbers = (56, 38, 41, 32, 45, 54, 31, 87, 58, 96) Partition(numbers, 0, 6) is called. Assume quicksort always chooses the element at the midpoint as the pivot.
What is the pivot?
What is the low partition?
What is the high partition?
What is numbers after Partition(numbers, 0, 6) completes?
The pivot for the Partition(numbers, 0, 6) operation is 54. The low partition consists of 56, 38, 41, 32, 45, and 31. The high partition consists 87, 58, and 96. After the completion of Partition(numbers, 0, 6), the updated order of the numbers would be (56, 38, 41, 32, 45, 31, 54, 87, 58, 96).
In the given scenario, the quicksort algorithm chooses the element at the midpoint (index 5) of the given range (from 0 to 6) as the pivot. In this case, the element at index 5 is 54. The Partition operation then divides the numbers into two partitions based on their relationship to the pivot. All elements less than the pivot are placed in the low partition, while all elements greater than the pivot are placed in the high partition.
In the given numbers, the elements less than 54 are 56, 38, 41, 32, 45, and 31, which form the low partition. The elements greater than 54 are 87, 58, and 96, which form the high partition. After the completion of the Partition operation, the numbers are rearranged so that the low partition comes before the pivot and the high partition comes after the pivot. Therefore, the updated order of the numbers becomes (56, 38, 41, 32, 45, 31, 54, 87, 58, 96).
Learn more about numbers here: brainly.com/question/13045235
#SPJ11
If a beta value has a large standard error, what can we conclude?
That estimates of beta vary little across different samples.
That estimates of beta vary widely across different samples.
The sampling distribution of beta is narrow
The estimate of beta in our sample is bigger than most other samples.
The larger the standard error, the less reliable the sample mean is as an estimate of the population mean. So, if a beta value has a large standard error, we can conclude that estimates of beta vary widely across different samples.
If a beta value has a large standard error, we can conclude that estimates of beta vary widely across different samples.What is a Beta Value?A beta value is a numerical value that represents the extent to which a variable affects another variable in a regression analysis. It's a measure of the sensitivity of a dependent variable to changes in an independent variable. A high beta value means that the variable has a significant impact on the dependent variable, whereas a low beta value indicates a lesser effect.What is Standard Error?The standard error of a statistic is an estimate of how much sampling variability it has. In other words, it is the standard deviation of its sampling distribution. It aids in determining the precision of the sample mean as an estimate of the population mean. The larger the standard error, the less reliable the sample mean is as an estimate of the population mean. So, if a beta value has a large standard error, we can conclude that estimates of beta vary widely across different samples.
To know more about standard visit:
https://brainly.com/question/31828911
#SPJ11
