simple transformatoin question of bernoulli distribution - Gambling and Probability

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simple transformatoin question of bernoulli distribution

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10-20-2012 , 07:01 PM

mugatu668

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Assume the random variable X has the Bernoulli distribution (i.e., binomial with n = 1): P(X = 1) = θ = 1 – P(X = 0).
(a) Find E(sqrt(x))

Okay, so...

E(x) would be the sum from 0 to infinity of

x*(θ^x)

(I think?? actually I'm getting myself confused thinking about this now)

So to do a transformation we should do

y = sqrt(x)
x = y^2

replace the x's with that, and then change bounds

But wait..wouldn't the bounds be sqrt(0) = 0 and sqrt(infinity) = infinity?

Umm...mabye I don't really understand the bernoulli distribution

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10-21-2012 , 02:00 AM

BruceZ

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You're making it way too complicated. x takes on the value of 1 with probability θ, and 0 with probability 1 - θ, so sqrt(x) takes on those same values with those same probabilities, and the expected value is just

1*θ + 0*(1-θ) = θ

The Bernoulli is the simplest distribution there is.

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10-21-2012 , 11:13 AM

mugatu668

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I see . So it's

sqrt(1)*theta + sqrt(0)(1-theta) just to be clear, right?

thanks for your help

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10-21-2012 , 02:10 PM

BruceZ

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Quote:

Originally Posted by mugatu668

I see . So it's

sqrt(1)*theta + sqrt(0)(1-theta) just to be clear, right?

Yes.

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10-21-2012 , 03:21 PM

mugatu668

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thanks!

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10-22-2012 , 12:58 AM

mugatu668

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Hmm this seems too simple though. It then asks me to find E(x^3) and E(x^10)

is this really so trivial that they all equal each other? maybe he is trying to drive home some point?

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10-22-2012 , 01:08 AM

BruceZ

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Quote:

Originally Posted by mugatu668

Hmm this seems too simple though. It then asks me to find E(x^3) and E(x^10)

is this really so trivial that they all equal each other? maybe he is trying to drive home some point?

Yes, those are all the same theta. You know the definition of expected value right? You take each possible value multiplied by the probability that value occurs, and sum those together.

Perhaps his point is that all of the moments are theta, and that will allow you to get any moment of a binomial distribution as n*theta since any moment of a binomial will be a sum of the moments of bernoulli trials because the expected values add.

Last edited by BruceZ; 10-22-2012 at 01:19 AM.

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10-22-2012 , 10:32 AM

reno expat

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Quote:

Originally Posted by mugatu668

Hmm this seems too simple though. It then asks me to find E(x^3) and E(x^10)

is this really so trivial that they all equal each other? maybe he is trying to drive home some point?

I'm assuming the point he is trying to drive home is LOTUS (law of the unconscious statistician), which is if you know the pdf (or pmf in the discrete case) of a random variable X, then E[g(X)] = g(X)*pdf(X)

in the case of the bernoulli distribution, the pmf is just (1-p) when X=0 and p when X=1

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10-22-2012 , 08:08 PM

reno expat

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Originally Posted by reno expat

slight mistake here. its E[g(X)] = integral over the support of X of g(X)*pdf(X) and in the case of a discrete distribution, the sum over the support of the PMF

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10-26-2012 , 04:24 AM

#10

Siegmund

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He may also be having you find E(X^2), E(X^3), etc (all the same, = theta), in preparation for having you stick those numbers into formulas like

Var(X) = E(X^2)-(E(X)^2) = theta - theta^2

and the similar formulas that exist for skewness, kurtosis, etc. If someone asks you "hey, here's a distribution, what's it's skewness?" you very rarely actually calculate E((X-mu)^3) directly, but rather look at the first 3 moments and combine them.

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