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xxTIMURxx [149]
3 years ago
15

There is a circle with a center of 0,0 on a coordinate plane. There is one point on the circle's circumference in which the x:y

ratio is 3:1. What is a possible coordinate?
Mathematics
1 answer:
abruzzese [7]3 years ago
6 0

Answer:

Step-by-step explanation:

Suppose the radius is 1. The parametric equations for the circle are

x = cosθ

y = sinθ

x:y = 3:1

tanθ = ⅓

cosθ = 3/√(1²+3²) =  3/√10

sinθ = 1/√10

The solutions are (3/√10, 1/√10) and (-3/√10, -1/√10).

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5.2.14. For the negative binomial pdf p (k; p, r) = k+r−1 (1 − p)kpr, find the maximum likelihood k estimator for p if r is know
Volgvan

Answer:

\hat p = \frac{r}{\bar x +r}

Step-by-step explanation:

A negative binomial random variable "is the number X of repeated trials to produce r successes in a negative binomial experiment. The probability distribution of a negative binomial random variable is called a negative binomial distribution, this distribution is known as the Pascal distribution".

And the probability mass function is given by:

P(X=x) = (x+r-1 C k)p^r (1-p)^{x}

Where r represent the number successes after the k failures and p is the probability of a success on any given trial.

Solution to the problem

For this case the likehoof function is given by:

L(\theta , x_i) = \prod_{i=1}^n f(\theta ,x_i)

If we replace the mass function we got:

L(p, x_i) = \prod_{i=1}^n (x_i +r-1 C k) p^r (1-p)^{x_i}

When we take the derivate of the likehood function we got:

l(p,x_i) = \sum_{i=1}^n [log (x_i +r-1 C k) + r log(p) + x_i log(1-p)]

And in order to estimate the likehood estimator for p we need to take the derivate from the last expression and we got:

\frac{dl(p,x_i)}{dp} = \sum_{i=1}^n \frac{r}{p} -\frac{x_i}{1-p}

And we can separete the sum and we got:

\frac{dl(p,x_i)}{dp} = \sum_{i=1}^n \frac{r}{p} -\sum_{i=1}^n \frac{x_i}{1-p}

Now we need to find the critical point setting equal to zero this derivate and we got:

\frac{dl(p,x_i)}{dp} = \sum_{i=1}^n \frac{r}{p} -\sum_{i=1}^n \frac{x_i}{1-p}=0

\sum_{i=1}^n \frac{r}{p} =\sum_{i=1}^n \frac{x_i}{1-p}

For the left and right part of the expression we just have this using the properties for a sum and taking in count that p is a fixed value:

\frac{nr}{p}= \frac{\sum_{i=1}^n x_i}{1-p}

Now we need to solve the value of \hat p from the last equation like this:

nr(1-p) = p \sum_{i=1}^n x_i

nr -nrp =p \sum_{i=1}^n x_i

p \sum_{i=1}^n x_i +nrp = nr

p[\sum_{i=1}^n x_i +nr]= nr

And if we solve for \hat p we got:

\hat p = \frac{nr}{\sum_{i=1}^n x_i +nr}

And if we divide numerator and denominator by n we got:

\hat p = \frac{r}{\bar x +r}

Since \bar x = \frac{\sum_{i=1}^n x_i}{n}

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