How do you multiply (x+5)(x+5)(x+5)

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

4 0

3 years ago

What would be a line perpendicular to (2,3) (5,5)?

ladessa [460]

9514 1404 393

Answer:

3x + 2y = 0 or y = -3/2x

Step-by-step explanation:

The relationship between a line and its perpendicular is that their slopes are opposite reciprocals of each other. So, the thing you need to know about the line through the two given points is its slope. That is found using the slope formula:

m = (y2 -y1)/(x2 -x1) . . . . . . where the two give points are (x1, y1) and (x2, y2)

Putting your point coordinates in this formula gives ...

m = (5 -3)/(5 -2) = 2/3

Then the perpendicular line will have a slope that is the opposite reciprocal:

slope of perpendicular = -1/m = -1/(2/3) = -3/2

You have not said whether the perpendicular line must meet any other requirement, so the simplest one we can write is ...

y = mx + b . . . . . m is the slope and b is the y-intercept

y = -3/2x . . . . . . has a y-intercept of 0 (goes through the origin)

Multiplying by 2 and adding 3x gives ...

3x + 2y = 0 . . . . . standard form equation

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<em>Comment on the attachment</em>

The geometry program I used to create the attachment wrote the equation of the line through the given points as ...

2x -3y = -5 . . . . . standard form equation

The equation of a perpendicular line can be created from this form by swapping the x- and y-coefficients and negating one of them. Simply swapping the coefficients would give ...

-3x +2y = <something>

In standard form, we want the leading coefficient to be positive, so we can choose to negate the x-coefficient. The <something> can be zero to make the line go through the origin.

3x +2y = 0 . . . . . a perpendicular in standard form

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If you need to have the line go through a particular point, you can determine the constant value (on the right of the = sign) by putting the point coordinates in for x and y.

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Method summary:

find the slope of the given line
determine its opposite reciprocal to find the slope of the perpendicular line
use that new slope in an appropriate form to have the line go where you want it to go. (Point-Slope Form is often useful)