Bernoulli sample question

[moogle12]

'Given the random sample {Xi} N i=1, with each Xi ∼ Bernoulli(1/2), what is the expectation of the sample mean when N = 10? What is the variance of the sample mean when N = 10? How do your answers change when N = 100? When N = 1, 000, 000? Explain your answers in reference to the Law of Large Numbers.'

I want to see where I stand with this and if you can point out any mistakes I make, so I can see if I'm on the right track with this. If I've done this correctly, when N=10, the expectation of the sample mean is 0.5 and the variance is zero. This immediately seems off to me, as the variance should decrease as the sample size increases, right? But if when the sample is very large it approximates a standard normal distribution, then the variance should be 1 when very large. It therefore seems implausible that the variance could be zero now. Can someone explain?

Thanks

 

[Romsek]

[MATH]E[X_i] = p = \dfrac 1 2[/MATH]
[MATH]E\left[\dfrac 1 N \sum \limits_{i=1}^N ~X_i \right]= \dfrac 1 N \sum \limits_{i=1}^N ~E[X_i] =\dfrac{Np}{N} = p = \dfrac 1 2[/MATH]
[MATH]Var[X_i] = E[X_i^2]-(E[X_i])^2 = (1^2 p + 0^2 (1-p)) - p^2 = p - p^2 = p(1-p)[/MATH]
When summing independent random variables variances add. When scaling random variables by a constant the variance is scaled by the constant squared.

[MATH]Var\left[\dfrac 1 N \sum \limits_{i=1}^N ~X_i\right] = \dfrac{1}{N^2} \sum \limits_{i=1}^N ~Var[X_i] = \dfrac{p(1-p)}{N} = \dfrac{1}{4N}[/MATH]
Now see if you can address your questions.

 

[moogle12]

'Given the random sample {Xi} N i=1, with each Xi ∼ Bernoulli(1/2), what is the expectation of the sample mean when N = 10? What is the variance of the sample mean when N = 10? How do your answers change when N = 100? When N = 1, 000, 000? Explain your answers in reference to the Law of Large Numbers.'

I want to see where I stand with this and if you can point out any mistakes I make, so I can see if I'm on the right track with this. If I've done this correctly, when N=10, the expectation of the sample mean is 0.5 and the variance is zero. This immediately seems off to me, as the variance should decrease as the sample size increases, right? But if when the sample is very large it approximates a standard normal distribution, then the variance should be 1 when very large. It therefore seems implausible that the variance could be zero now. Can someone explain?

Thanks

[MATH]E[X_i] = p = \dfrac 1 2[/MATH]
[MATH]E\left[\dfrac 1 N \sum \limits_{i=1}^N ~X_i \right]= \dfrac 1 N \sum \limits_{i=1}^N ~E[X_i] =\dfrac{Np}{N} = p = \dfrac 1 2[/MATH]
[MATH]Var[X_i] = E[X_i^2]-(E[X_i])^2 = (1^2 p + 0^2 (1-p)) - p^2 = p - p^2 = p(1-p)[/MATH]
When summing independent random variables variances add. When scaling random variables by a constant the variance is scaled by the constant squared.

[MATH]Var\left[\dfrac 1 N \sum \limits_{i=1}^N ~X_i\right] = \dfrac{1}{N^2} \sum \limits_{i=1}^N ~Var[X_i] = \dfrac{p(1-p)}{N} = \dfrac{1}{4N}[/MATH]
Now see if you can address your questions.

Thanks! Using your method I see that the variance when N=10 is 0.025. When I got a variance of zero, I used the definition of the variance

[MATH]E[X_i] = p = \dfrac 1 2[/MATH]
[MATH]E\left[\dfrac 1 N \sum \limits_{i=1}^N ~X_i \right]= \dfrac 1 N \sum \limits_{i=1}^N ~E[X_i] =\dfrac{Np}{N} = p = \dfrac 1 2[/MATH]
[MATH]Var[X_i] = E[X_i^2]-(E[X_i])^2 = (1^2 p + 0^2 (1-p)) - p^2 = p - p^2 = p(1-p)[/MATH]
When summing independent random variables variances add. When scaling random variables by a constant the variance is scaled by the constant squared.

[MATH]Var\left[\dfrac 1 N \sum \limits_{i=1}^N ~X_i\right] = \dfrac{1}{N^2} \sum \limits_{i=1}^N ~Var[X_i] = \dfrac{p(1-p)}{N} = \dfrac{1}{4N}[/MATH]
Now see if you can address your questions.

Thanks! I realise exactly what I did wrong now. Your help is appreciated.