Finding moment generating functions

[DODGEVIPER13]

Homework Statement

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Homework Equations

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The Attempt at a Solution

Well I believe it to be Binomial as the probabilities never change. So I used it to solve for the moment generating function, I do howvere want to know if what I have done is correct and also how I went from sigma((Pe^t)^y(n y)q^(n-y)) as the book did this?

 

[haruspex]

Isn't p the probability of one head in one toss? How is it 0.75?

 

[DODGEVIPER13]

Ok so p should be .50 but does my work look ok other than that and the obvious mathematical error due to that on the second part

 

[DODGEVIPER13]

After the final sigma part like I asked above how do I arrive at the moment generating function like I did?

 

[haruspex]

In the third line of part 2, you've written E[Y] = M'(t) by mistake for M'(0).
In the next line, you've left something out when applying the product rule to get M''.

 

[haruspex]

After the final sigma part like I asked above how do I arrive at the moment generating function like I did?

It's just the binomial expansion (run in reverse).

 

[DODGEVIPER13]

Ok a bit lost on what was wrong with m" it looks right to me hmmm where should I look

 

[haruspex]

Ok a bit lost on what was wrong with m" it looks right to me hmmm where should I look

Looks like I need to take a break. You're quite right, and the rest of your work looks good too.

 

[DODGEVIPER13]

Ok so should I change p to .5 and q to .5 and I wind up with 1.5

 

[DODGEVIPER13]

I changed p from .75 to .5 as you said

 

[DODGEVIPER13]

Also for the first part I now have (.50e^t+.50)^2

 

[haruspex]

All good.

 

[DODGEVIPER13]

Wait why is p what you said should I not also include the probability that you flip a head and head thus making it .75

 

[DODGEVIPER13]

When you say all good was it all good before or all good now?

 

[haruspex]

When you wrote the first two lines of your solution, what were you defining p to be? What do you get if you set n=2 and write out the sum in the second line?
You showed E[Y] = np. If n=2 and p=.75, would you really expect an average of 1.5 heads in two tosses?

 

[DODGEVIPER13]

I was defining p to be the possible heads observed by the 2 flips and to the final question no I would not so I guess what I have changed it to is correct

 

[haruspex]

I was defining p to be the possible heads observed by the 2 flips

But p is a probability, not a number of heads. If you wanted to approach it that way you'd need three probabilities: 0, 1 or 2 heads.

 

[DODGEVIPER13]

Ah ok got it now thanks so the way I have it now is correct

 

[haruspex]

Ah ok got it now thanks so the way I have it now is correct

Yes.