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This is for beginners in probability!

Could someone give me a step by step on how to find the MGF of the binomial distribution?

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    Ok, I feel kinda bad for my answer (although this question deserves it), so I'll give a real hint: The binomial distribution can be thought of as a sum of independent, identically distributed (IID) random variables. The MGF of a sum of IID variables has a simple expression in terms of the MGF of the parent distribution. The MGF for the parent distribution is simple in this case. – 2017-01-23
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    @spaceisdarkgreen please realize that this stackexchange is open to everyone, not just hardcore mathematicians. Just trying to give a helping hand to those who are getting started :) – 2017-01-23
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    I wasn't responding to the easiness of the question, but to the fact that you asked without it without any context or indication of what difficulties you were having with the problem. (And I'm by no means a hardcore mathematician.) – 2017-01-23
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    Okay I just know so many students who get discouraged on these exchanges because they need help with the basics first :) I saw that the forum is lacking in beginner material that would be super help to students and thought I could help with that :) – 2017-01-23
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    Gotcha - asking questions that you know the answer to just get it up on the website goes somewhat against the norms of what typically goes on here, although I don't think it's outright against the rules (answering your own question is certainly not discouraged in general.) But asking homework-type questions / requesting derivations *without giving any context or sense of where the difficulty lies and what approaches have been tried* is most certainly frowned upon. – 2017-01-23
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    I would mention in the body of the Question that MGF is an abbreviation for "moment generating function". – 2017-01-23
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    A better way to approach this, I think, would be to post: "I initially had difficulty deriving the MGF for the binomial distribution because (give your reasons). The derivation I came up with was: (your derivation). Is this the best way to do it? Could I have approached it differently? – 2017-01-23
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    @spaceisdarkgreen thank you so much for the tips! I'll phrase my questions better and make sure they are not homework type questions next time :) – 2017-01-23
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    @Phia-CM I think you misunderstood. It's totally ok if they're homework-type questions. What's not ok is not giving any context. The reason is without context it's hard to know where the asker is having trouble and thus to write a good answer that addresses the difficulties the asker is having at the right level. Also, a lot of people come on here asking for homework answers but being completely unwilling to try to understand, so it helps to avoid that too. – 2017-01-23

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  1. Write down the definition of a MGF.

  2. Write down the definition of the binomial distribution.

  3. Plug the binomial distribution into the definition of MGF.

  4. Clean up the expression.

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    So the reason why I posted this question and then the solution is because when I was starting out as a stats major, they assumed that the cleaning up expression would be super easy. There are a bunch of different ways and it can overwhelming. I wanted to give a nice explanation to those who are struggling :) – 2017-01-23
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    @Phia-CM If that was your motivation you should have indicated that in the question. Would have been more than willing to help. The easiest way is to calculate the MGF of the Bernoulli distribution and take it to the $n$-th power since the binomial is the sum of n independent Bernoullis – 2017-01-23
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First remember the definition of a MGF

$m_Y(t) = E[e^{tY}]$ For any random variable Y

so for a random variable Y with the binomial distribution

$m_Y(t)= \sum_{y=0}^n e^{ty}{n\choose y} p ^y q^{n-y} $

First step is to combined exponents

$m_Y(t)= \sum_{y=0}^n {n\choose y} (pe^{t})^y q^{n-y} $

The most important part here is to know that

$(x+y)^n = \sum_{i=0}^n {n\choose i}x^i y^{n-i}$

When looking at are sum we can see that $x = pe^{t}$ and $y = q$

So we can come to the conclusion that

$m_Y(t) = (pe^{t} + q)^n$