# How do you find the Probability Generating Function?

• Orphen89
In summary: Wow, this is so helpful! I definitely understand it now. Thank you so much for explaining it to me!In summary, we discussed a branching process with branching probabilities given by P0=1/2 and Pj=1/3^{j} for j \geq 1. We were asked to find the probability generating function, and after some calculations, we arrived at the answer G(x) = (x+3)/(2(3-x)) by using the formula \frac {1}{1 - \frac {x}{3}} = \frac {3}{3 - x } and subtracting the initial term of p0=1/2. We also discussed the mean and variance of a discrete probability distribution and
Orphen89

## Homework Statement

Consider a branching process with branching probabilities given by P0=1/2 and Pj=1/3$$^{j}$$ for j $$\geq$$ 1

Find the probability generating function: $$\sum^{\infty}_{n=0} p_{n}x^{n}$$

## The Attempt at a Solution

Now, the answer is supposed to be G(x) = (x+3)/(2(3-x)), but I really have no idea how to get this answer. I've checked it up on the internet and haven't found anything that looks even vaguely helpful, and since I'm only meeting my tutor later in the week I'm pretty much stuck on my own for this one. Is there some sort of formula you can use to find the probability generating function? If possible, could you please show me how to get that answer from the given data?

$$G(x) = \frac {1}{2} + \frac {1}{3} x + \frac {1}{9} x^{2} + \frac {1}{27} x^{3} + ...$$

$$1 + \frac {1}{3} x + \frac {1}{9} x^{2} + \frac {1}{27} x^{3} + ... = \frac {1}{1 - \frac {x}{3}} = \frac {3}{3 - x }$$

$$G(x) = \frac {3} {3-x} - \frac {1}{2} = \frac {x+3}{2(3-x)}$$

Thanks for the quick reply! I just have two questions: In the second line, is it supposed to be 1/2 or 1? And in the third line, you're basically just subtracting the p0 value from G(x) right?

Thanks for the help.

Orphen89 said:
Thanks for the quick reply! I just have two questions: In the second line, is it supposed to be 1/2 or 1? And in the third line, you're basically just subtracting the p0 value from G(x) right?

Thanks for the help.

The second line is a geometric series. But it's 1/2 more than what we want.

Last edited:
Sorry I just have one more question - does this mean that for p0= 1/3, p1=1/4, p2=1/6 and pn=1/2^n that the probability generating function is equal to -(7x-6)/(4(x-2)) + 1/3 + (1/4)x + (1/6)x^2?

Also, if it is incorrect, could someone please show me the correct method? For some reason I'm having a fair bit of difficulty getting my mind around this...

Are you saying that $$p_{n}= (\frac {1}{2})^{n}$$ for n > 2?

Then your answer is correct if $$\sum^{\infty}_{3}p_{n}x^{n} = -\frac {7x-6}{4(x-2)}$$

but that's not what I'm getting

Last edited:
Sorry about that! I forgot about that part - it is pn=1/2^n for n $$\geq$$ 3. Thanks for all the help Random Variable, I appreciate it (all this Generating Function stuff is going to be on an exam coming up in a few weeks...)

I'm getting $$\frac {x^{3}}{4(2-x)}$$

Really? I think you might be right, because I can remember hearing that that was the answer...

I originally thought that (using the data), I could create a probability generating function with a/(1-a) - 1/3 - 1/4 - 1/6, with a = x/2

Do you know where I have gone wrong, (or if I even have the right equation...?) >_<

see below

Last edited:
We know $$1 + \frac {1}{2} x + \frac {1}{4} x^{2} + \frac {1}{8} x^{3} + \frac {1}{16}x^{4} + ... = \frac {1}{1 - \frac {x}{2}}$$ (infinite geometric series with common ratio x/2 and first term of 1)

What we want is $$\frac {1}{8} x^{3} + \frac {1}{16} x^{4} + \frac {1}{32}x^5+ ...$$

so subtract the terms we don't want

$$\frac {1}{1 - \frac {x}{2}} - 1 - \frac {1}{2}x - \frac {1}{4}x^2$$

Last edited:
As a side note, to find the mean of a discrete probability distribution, take the derivative of the associated probability generating function and evaluate it at x=1.

Wow, I think I finally understand how to do it now. Thanks for all the help Random Variable, you managed to explain the probability generating function to me better than my tutor =D

And if you take the second derivative of a probability generating function and evaluate it at x=1, you'll get

$$E(X(X-1)) = E(X^{2}) - E(X) = E(X^{2}) - \mu$$

and since $$V(X) = E[(X- \mu)^{2}] = E(X^{2}) - \mu^2$$

$$V(X) = E(X(X-1)) + \mu -\mu^{2}$$

## 1. What is a Probability Generating Function?

A Probability Generating Function (PGF) is a mathematical tool used in probability theory to describe the probability distribution of a discrete random variable. It is a generating function that contains information about the probabilities of all possible values of the random variable.

## 2. How do you calculate a Probability Generating Function?

The formula for a Probability Generating Function is PGF(t) = E(t^X), where X is the random variable and t is the variable of the generating function. To calculate the PGF, you need to find the expected value of t^X for all possible values of X and plug them into the formula.

## 3. What is the purpose of using a Probability Generating Function?

The main purpose of a Probability Generating Function is to easily and efficiently calculate the probabilities of all possible values of a discrete random variable. It can also be used to determine moments, such as the mean and variance, of the random variable.

## 4. Can a Probability Generating Function be used for continuous random variables?

No, a Probability Generating Function is only applicable to discrete random variables. For continuous random variables, the equivalent function is called the Moment Generating Function (MGF).

## 5. What are the advantages of using a Probability Generating Function?

Using a Probability Generating Function can simplify complex probability calculations and provide a systematic way to determine the probabilities of all possible outcomes of a discrete random variable. It can also be used to easily determine moments of the random variable.

• Calculus and Beyond Homework Help
Replies
3
Views
727
• Calculus and Beyond Homework Help
Replies
2
Views
466
• Calculus and Beyond Homework Help
Replies
14
Views
498
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
686
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
967
• Calculus and Beyond Homework Help
Replies
4
Views
1K