Statistics(moment generating function)

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The discussion covers three statistical problems related to probability and moment-generating functions. For the first problem, the probability P(0.45 < X < 0.55) for a uniform distribution is calculated as 0.10, but clarification is provided that this pertains to a single value rather than the mean. The second problem involves deriving the moment-generating function for the normal distribution, with guidance on using transformations of random variables. The final question addresses finding a function whose variance is independent of the parameter p, suggesting the application of variance formulas for transformed normal random variables. The participants emphasize the importance of understanding the central limit theorem and transformations in statistics.
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1. Let X denote the mean of a random sample of size 75 from the distribution that has the pdf f (x) =1, 0<=x<=1. Calculate P (0.45 <X< 0.55).
2. Derive the moment-generating function for the normal density.
3. Let Y n (or Y for simplicity) be b (n, p). Thus, Y / n is approximately N [p, p (1 -p) / n]. Statisticians often look for functions of statistics whose variances do not depend upon the parameter. Can you find a function, say u(Y / n), whose variance is essentially free of p?


can anyone help me with them?
 
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I'm not going to risk giving advise on your other 2 problems since I'm not entirely sure what they are looking for but I can give you help on #1.

When they say that a distribution has a Pdf of F(x)=1 they mean that it is http://en.wikipedia.org/wiki/Uniform_distribution_(continuous)" in other words all events within the defined space of 0<=x<=1 are equal. Therefore all you need to do to find the area (and thus probability) within two defined points (events) in this distribution is find the absolute value of their distance.

In the case of this problem: P(0.45<x<0.55)= 0.55-0.45 = 0.10

I hope that helps. ^-^
 
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thanks for your reply.
 
About #1, if the question is "the mean of a random sample of size 75 from the distribution that has the pdf f (x) =1" then the answer is not simply 0.1 - that is the probability associated with a single value of x, not the mean. Given the indicated level of the questions, perhaps think about the central limit theorem.

For the second problem, note that if X, Y are random variables, and

<br /> Y = \mu + \sigma X<br />

then the moment generating function of Y is

\begin{align*}<br /> \phi_Y(s) = E\left(e^{sY}\right) &amp;= \int e^{s(\mu + \sigma X)} f(x) \, dx \\<br /> &amp;= e^{s\mu} \int e^{(s \, \sigma ) X} f(x) \, dx \\<br /> &amp; = s^{s\mu} \phi_X (s \, \sigma)<br /> \end{align*}<br />

so if you know the mgf of X you can get the mgf using the idea in the final line. That should help you with the mgf for the normal distribution.

The final question is a classic question seen when transformations of random variables are introduced. Begin with the idea stated, that Y/n is roughly normal, with the given variance. You know the formula for the variance of a transformed normal r.v. - apply it to the given constant, set the result equal to a constant C, and solve the (very simple) differential equation.
 
statdad said:
About #1, if the question is "the mean of a random sample of size 75 from the distribution that has the pdf f (x) =1" then the answer is not simply 0.1 - that is the probability associated with a single value of x, not the mean. Given the indicated level of the questions, perhaps think about the central limit theorem.
You know what he is exactly right, I read that to calculate for a single value of X instead of for the mean X, I apologize. Thank you for catching my mistake statdad.
 
"You know what he is exactly right, I read that to calculate for a single value of X instead of for the mean X, I apologize. Thank you for catching my mistake statdad."

You are welcome, but there is no need for gnashing of teeth - we all make mistakes here, whether simple typing errors or others. I'm sure I'm far ahead of you in that regard.
I hope the OP is making progress on these.
 
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