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3 years ago

How do you know whether to quanities x and y are proportional

1 answer:

5 0

Ratios are proportional if they represent the same relationship. One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional.

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

4 0

3 years ago

8th term of 1/4,1/8,1/16,1/32

sukhopar [10]

$a_1=\dfrac{1}{4};\ a_2=\dfrac{1}{8};\ a_3=\dfrac{1}{16};\ a_4=\dfrac{1}{32};...\\\\it's\ a\ geometric\ sequence\ where\ a_1=\dfrac{1}{4}\ and\ r=\dfrac{1}{8}:\dfrac{1}{4}=\dfrac{1}{2}\\\\a_n=a_1r^{n-1}\\\\n=8\\\\a_8=\dfrac{1}{4}\cdot\left(\dfrac{1}{2}\right)^{7-1}=\dfrac{1}{4}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{4}\cdot\dfrac{1}{64}=\dfrac{1}{256}$

4 0

3 years ago

The opposite of the fraction one third

lozanna [386]

The opposite of the fraction one third would be negative one third. To be opposite, they must have differing signs. One number should be positive and the other number should be negative. It is different from reciprocal. To be a reciprocal, <span>one number should be the flipped fraction, or upside down version, of the other number.</span>

3 0

3 years ago

Graph the function<br> Need help ASAP plz

sattari [20]

Given:

The equation of a function is

$y=(x-4)^2+1$