Expand and simplify 2(2x+7) + 3(x+ 1)

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

Please help me, i need the answers ASAP! (With proper steps please)

docker41 [41]

a) 35 x 13 + 3.14 x $\frac{13}{2}$ ²

= 455 + 3.14 x 42.25

= 455 + 132.665

= 587.665 m²

∴ The area is 587.665 m².

4 0

3 years ago

When adding two rational

nasty-shy [4]

Answer:

Obviously!! positive is your answer. thanks!!

6 0

3 years ago

Which of the following is not a prime factor?

Liono4ka [1.6K]

63 is the answer to your problem. Hope this helps!

8 0

3 years ago

Read 2 more answers

solve for each please i really need help if u want to help me with my test and i get an a or a b i will give u 500 dollars

Len [333]

Answer:

The relation is not a function

The domain is {1, 2, 3}

The range is {3, 4, 5}

Step-by-step explanation:

A relation of a set of ordered pairs x and y is a function if

Every x has only one value of y

x appears once in ordered pairs

<u><em>Examples:</em></u>

The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)

The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)

The domain is the set of values of x

The range is the set of values of y

Let us solve the question

∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}

∵ x = 1 has y = 3

∵ x = 2 has y = 3

∵ x = 3 has y = 4

∵ x = 2 has y = 5

→ One x appears twice in the ordered pairs

∵ x = 2 has y = 3 and 5

∴ The relation is not a function because one x has two values of y

∵ The domain is the set of values of x

∴ The domain = {1, 2, 3}

∵ The range is the set of values of y

∴ The range = {3, 4, 5}

3 0

3 years ago