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3 years ago

(−4 + 5) − (6 + 7) = x x 0

Mathematics

1 answer:

weeeeeb [17]3 years ago

3 0

Add the numbers

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−

)

−

(

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({\color{#c92786}{-4}}+{\color{#c92786}{5}})-(6+7)=xx^{0}

(−4+5)−(6+7)=xx0

(

)

−

(

)

({\color{#c92786}{1}})-(6+7)=xx^{0}

(1)−(6+7)=xx0

Add the numbers

−

(

)

1-({\color{#c92786}{6}}+{\color{#c92786}{7}})=xx^{0}

1−(6+7)=xx0

−

(

)

1-({\color{#c92786}{13}})=xx^{0}

1−(13)=xx0

Multiply the numbers

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⋅

1{\color{#c92786}{-1}} \cdot {\color{#c92786}{13}}=xx^{0}

1−1⋅13=xx0

−

1{\color{#c92786}{-13}}=xx^{0}

1−13=xx0

Subtract the numbers

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{\color{#c92786}{1-13}}=xx^{0}

1−13=xx0

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{\color{#c92786}{-12}}=xx^{0}

−12=xx0

Combine exponents

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-12={\color{#c92786}{xx^{0}}}

−12=xx0

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-12={\color{#c92786}{x^{1}}}

−12=x1

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Solution

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find the present value of an investment that is worth $19,513.75 after earning 3percent simple interest for 5.5 years

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Answer:

present value = $16750

Step-by-step explanation:

The simple interest formula allows us to calculate A, which is the final amount. According to this formula, the amount is given by A = P (1 + r*t), where P is the principal, r is the annual interest rate in decimal form, and t is the loan period expressed in years

simple interest formula:

t: time

P: present value

A: amount

r : anual interest

A = P (1 + r*t)

P = A / (1 + r*t)

P = 19,513.75 / (1 + 3/100 * 5.5)

P = 19,513.75/ (1 + 0.165)

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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 $

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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:

$\hat a = max(X_i)$

$E(a) - a= 0$ 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.

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