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love history [14]
3 years ago
5

Let X1 and X2 be independent random variables with mean μand variance σ².

Mathematics
1 answer:
My name is Ann [436]3 years ago
7 0

Answer:

a) E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu

So then we conclude that \hat \theta_1 is an unbiased estimator of \mu

E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu

So then we conclude that \hat \theta_2 is an unbiased estimator of \mu

b) Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}

Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}

Step-by-step explanation:

For this case we know that we have two random variables:

X_1 , X_2 both with mean \mu = \mu and variance \sigma^2

And we define the following estimators:

\hat \theta_1 = \frac{X_1 + X_2}{2}

\hat \theta_2 = \frac{X_1 + 3X_2}{4}

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

E(\hat \theta_i) = \mu , i = 1,2

So let's find the expected values for each estimator:

E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})

Using properties of expected value we have this:

E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu

So then we conclude that \hat \theta_1 is an unbiased estimator of \mu

For the second estimator we have:

E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})

Using properties of expected value we have this:

E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu

So then we conclude that \hat \theta_2 is an unbiased estimator of \mu

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

Var(aX) = a^2 Var(X)

For the first estimator we have:

Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})

Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]

Since both random variables are independent we know that Cov(X_1, X_2 ) = 0 so then we have:

Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}

For the second estimator we have:

Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})

Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]

Since both random variables are independent we know that Cov(X_1, X_2 ) = 0 so then we have:

Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}

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Answer:

43.75 ft²

Step-by-step explanation:

s_l = (l√(w/2)² + h²) + (w√(l/2)² + h²)

l & w become 3.5, and h becomes 6.

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<em>Step 1:Because this is a square pyramid, what you see above essentially becomes what you see below.</em>

<em />s_l = 2(3.5√(3.5/2)² + 6²)

<em>Step 2: Divide 3.5 by 2 to get 1.75.</em>

<em />s_l<em> </em>= 2(3.5√1.75² + 6²)

<em>Step 3: Square both 1.75 and 6 to get 3.0625 and 36 respectively.</em>

s_l = 2(3.5√3.0625 + 36)

<em>Step 4: Add 3.0625 and 36 to get 39.0625.</em>

<em />s_l = 2(3.5√39.0625)

<em>Step 5: The square root of 39.0625 is 6.25.</em>

<em />s_l<em> </em>= 2(3.5 * 6.25)

<em>Step 6: Multiply 3.5 by 6.25 to get 21.875.</em>

<em />s_l = 2(21.875)

<em>Step 7: Multiply 2 by 21.875 to get 43.75.</em>

<em />s_l = 43.75 ft²

The lateral area of this pyramid is 43.75 ft².

<em />

<em />

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