Slove and show working for 1/4 ÷ 3

Show that if X is a geometric random variable with parameter p, then

Answer:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

In order to find the expected value E(1/X) we need to find this sum:

$E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}$

Lets consider the following series:

$\sum_{k=1}^{\infty} b^{k-1}$

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

$\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}$ (a)

On the last step we assume that $0\leq r\leq b$ and $\sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}$ , then the integral on the left part of equation (a) would be 1. And we have:

$\int_{0}^b \frac{1}{1-r}dr=-ln(1-b)$

And for the next step we have:

$\sum_{k=1}^{\infty} \frac{b^{k-1}}{k}=\frac{1}{b}\sum_{k=1}^{\infty}\frac{b^k}{k}=-\frac{ln(1-b)}{b}$

And with this we have the requiered proof.

And since $b=1-p$ we have that:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

4 0

3 years ago

Which of the following are written in point-slope

tino4ka555 [31]

If I get a thanks I’ll answer on a diff comment

7 0

2 years ago

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A cylindrical swimming pool has a diameter of 14 feet and a height of 5 feet. How many gallons of water can the pool contain? Ro

Vinvika [58]

The volume of a cylinder is pi*height*radius^2 = pi*5*(14/2)^2 = 769.69 ft^3. To get the total gallons multiply 769.69 ft^3 by 7.5gal/ft^3 to get 5772.6765 gallons or 5773 gallons rounded to nearest whole number.

8 0

3 years ago

Jack spends 1 1/2 as long on his homework as Jill. Last week Jill spent 9 1/2 hours doing homework. How long did Jack spend doin

alina1380 [7]

The amount of time Jill spent to do her homework
= 9 1/2 hours. which equals to 570 minutes.

Which means, the amount of times Jack needed to do his is
= 570 minutes x 1.5
= 855 minutes which equals to 14 1/4 hours

hope this helps

7 0

3 years ago

A bag contains times with the letters P-R-O-B-A-B-I-L-I-T-Y. Tyler chooses a tire without looking and doesn't replace it. He cho

Yuri [45]

2/11 because there is 11 letters which would be the denominator and then there is 2 I's which would be the numerator

4 0

3 years ago

Read 2 more answers