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Math  /  Data & Statistics

Question22. Let XX denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of XX is f(x,θ)={(θ+1)xθ0≤x≤10 otherwise f(x, \theta)=\left\{\begin{array}{cl} (\theta+1) x^{\theta} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. where −1<θ-1<\theta. A random sample of ten students yields data x1=.92,x2=.79,x3=.90,x4=.65x_{1}=.92, x_{2}=.79, x_{3}=.90, x_{4}=.65, x5=.86,x6=.47,x7=.73,x8=.97,x9=.94x_{5}=.86, x_{6}=.47, x_{7}=.73, x_{8}=.97, x_{9}=.94, x10=.77x_{10}=.77. a. Use the method of moments to obtain an estimator of θ\theta and then compute the estimate for this data. b. Obtain the maximum likelihood estimator of θ\theta and then compute the estimate for the given data.

Studdy Solution

STEP 1

1. The probability density function (pdf) given is valid for 0≤x≤10 \leq x \leq 1.
2. The method of moments involves equating sample moments to population moments.
3. The maximum likelihood estimator (MLE) involves maximizing the likelihood function.

STEP 2

1. Use the method of moments to find an estimator for θ\theta.
2. Compute the estimate of θ\theta using the sample data.
3. Derive the maximum likelihood estimator for θ\theta.
4. Compute the MLE estimate of θ\theta using the sample data.

STEP 3

Calculate the expected value of XX using the pdf:
E[X]=∫01x⋅(θ+1)xθ dx E[X] = \int_0^1 x \cdot (\theta + 1) x^\theta \, dx
=(θ+1)∫01xθ+1 dx = (\theta + 1) \int_0^1 x^{\theta + 1} \, dx
=(θ+1)[xθ+2θ+2]01 = (\theta + 1) \left[ \frac{x^{\theta + 2}}{\theta + 2} \right]_0^1
=θ+1θ+2 = \frac{\theta + 1}{\theta + 2}

STEP 4

Set the sample mean equal to the expected value:
xˉ=θ+1θ+2 \bar{x} = \frac{\theta + 1}{\theta + 2}
Calculate the sample mean:
xˉ=0.92+0.79+0.90+0.65+0.86+0.47+0.73+0.97+0.94+0.7710 \bar{x} = \frac{0.92 + 0.79 + 0.90 + 0.65 + 0.86 + 0.47 + 0.73 + 0.97 + 0.94 + 0.77}{10}
xˉ=8.0010=0.80 \bar{x} = \frac{8.00}{10} = 0.80
Solve for θ\theta:
0.80=θ+1θ+2 0.80 = \frac{\theta + 1}{\theta + 2}
0.80(θ+2)=θ+1 0.80(\theta + 2) = \theta + 1
0.80θ+1.60=θ+1 0.80\theta + 1.60 = \theta + 1
0.20θ=0.60 0.20\theta = 0.60
θ=3 \theta = 3

STEP 5

Write the likelihood function:
L(θ)=∏i=110f(xi,θ)=∏i=110(θ+1)xiθ L(\theta) = \prod_{i=1}^{10} f(x_i, \theta) = \prod_{i=1}^{10} (\theta + 1) x_i^\theta
=(θ+1)10∏i=110xiθ = (\theta + 1)^{10} \prod_{i=1}^{10} x_i^\theta
Take the natural logarithm of the likelihood function:
ln⁡L(θ)=10ln⁡(θ+1)+θ∑i=110ln⁡xi \ln L(\theta) = 10 \ln(\theta + 1) + \theta \sum_{i=1}^{10} \ln x_i
Differentiate with respect to θ\theta and set the derivative to zero:
ddθln⁡L(θ)=10θ+1+∑i=110ln⁡xi=0 \frac{d}{d\theta} \ln L(\theta) = \frac{10}{\theta + 1} + \sum_{i=1}^{10} \ln x_i = 0

STEP 6

Solve for θ\theta:
10θ+1=−∑i=110ln⁡xi \frac{10}{\theta + 1} = -\sum_{i=1}^{10} \ln x_i
θ+1=−10∑i=110ln⁡xi \theta + 1 = -\frac{10}{\sum_{i=1}^{10} \ln x_i}
θ=−10∑i=110ln⁡xi−1 \theta = -\frac{10}{\sum_{i=1}^{10} \ln x_i} - 1
Calculate ∑i=110ln⁡xi\sum_{i=1}^{10} \ln x_i:
∑i=110ln⁡xi=ln⁡(0.92)+ln⁡(0.79)+ln⁡(0.90)+ln⁡(0.65)+ln⁡(0.86)+ln⁡(0.47)+ln⁡(0.73)+ln⁡(0.97)+ln⁡(0.94)+ln⁡(0.77) \sum_{i=1}^{10} \ln x_i = \ln(0.92) + \ln(0.79) + \ln(0.90) + \ln(0.65) + \ln(0.86) + \ln(0.47) + \ln(0.73) + \ln(0.97) + \ln(0.94) + \ln(0.77)
≈−0.0834−0.2363−0.1054−0.4308−0.1508−0.7550−0.3147−0.0305−0.0619−0.2614 \approx -0.0834 -0.2363 -0.1054 -0.4308 -0.1508 -0.7550 -0.3147 -0.0305 -0.0619 -0.2614
≈−2.4302 \approx -2.4302
Substitute back to find θ\theta:
θ=−10−2.4302−1 \theta = -\frac{10}{-2.4302} - 1
θ≈3.11 \theta \approx 3.11
The method of moments estimate of θ\theta is 3\boxed{3}.
The maximum likelihood estimate of θ\theta is 3.11\boxed{3.11}.

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