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Prove the central limit theorem for a sequence of i.i.d. Bernoulli($p$) random variables
$begingroup$
Prove the central limit theorem for a sequence of i.i.d.
Bernoulli($p$) random variables, where $pin(0,1)$.
I am trying to do this by computing the moment generating function of the object I want the limit of and use Taylor's expansion to show that it converges to the moment generating function of a standard normal.
Attempt:
For a random variable X, we have the moment generating function $$M_{X}(t)=mathbb{E}[e^{tX}]$$ and if we expand using the Taylor series of $e^{tX}$ we get $$M_X(t)=sum_{n=0}^infty frac{mathbb{E}[X^n]}{n!}t^n.$$ So, we have in particular $$M_x^{(n)}(0)=mathbb{E}[X^n].$$
Now for the proof, we have a Bernoulli random variable is 1 with probability $p$ and 0 with probability $(1-p)$. First we want the moment generating function for a Bernoulli random variable. In particular, we have $$M_{Bernoulli(p)}=(1-p)+pe^t=1+(e^t-1)p$$
Then, we have a binomial random variable as the sum of n independent Bernoulli variables. So,
$$M_{binomial(n,p)}=((1-p)+pe^t=(1+(e^t-1)p)^n$$
Suppose that $p=lambda/n$ and notice that $$M_{binomial(n,lambda/n)}=bigg(1+frac{(e^t-1)lambda}{n}bigg)^n rightarrow e^{lambda(e^t-1)}$$
I am not sure how to finish the proof though or if I am even on the right track...
probability probability-theory random-variables central-limit-theorem
asked Feb 2 at 1:18
MathIsHardMathIsHard
1,318516
$endgroup$
add a comment |
$begingroup$
Prove the central limit theorem for a sequence of i.i.d.
Bernoulli($p$) random variables, where $pin(0,1)$.
I am trying to do this by computing the moment generating function of the object I want the limit of and use Taylor's expansion to show that it converges to the moment generating function of a standard normal.
Attempt:
For a random variable X, we have the moment generating function $$M_{X}(t)=mathbb{E}[e^{tX}]$$ and if we expand using the Taylor series of $e^{tX}$ we get $$M_X(t)=sum_{n=0}^infty frac{mathbb{E}[X^n]}{n!}t^n.$$ So, we have in particular $$M_x^{(n)}(0)=mathbb{E}[X^n].$$
Now for the proof, we have a Bernoulli random variable is 1 with probability $p$ and 0 with probability $(1-p)$. First we want the moment generating function for a Bernoulli random variable. In particular, we have $$M_{Bernoulli(p)}=(1-p)+pe^t=1+(e^t-1)p$$
Then, we have a binomial random variable as the sum of n independent Bernoulli variables. So,
$$M_{binomial(n,p)}=((1-p)+pe^t=(1+(e^t-1)p)^n$$
Suppose that $p=lambda/n$ and notice that $$M_{binomial(n,lambda/n)}=bigg(1+frac{(e^t-1)lambda}{n}bigg)^n rightarrow e^{lambda(e^t-1)}$$
I am not sure how to finish the proof though or if I am even on the right track...
probability probability-theory random-variables central-limit-theorem
asked Feb 2 at 1:18
MathIsHardMathIsHard
1,318516
$endgroup$
1
$begingroup$
What you have done so far is merely deriving (correctly) Binomial becoming Poisson under the limit $n to infty, p to 0$ with $np = lambda$ held constant. The statement of Central Limit Theorem involves $frac{bar X - mu}{ sigma}$. Where's your analysis on $bar X$?
$endgroup$
– Lee David Chung Lin
Feb 2 at 2:05
$begingroup$
Oh okay. Thank you. So I want to find the mean and the std deviation for the random variable next?
$endgroup$
– MathIsHard
Feb 2 at 2:48
1
$begingroup$
There are many ways to do this. Since you wanted to use the moment generating function, you should find the MGF of $M_Y(t)$ where $Y equiv bar X = frac1n sum X_i$ and then further consider the MGF of $sqrt{n}frac{Y - mu}{ sigma}$
$endgroup$
– Lee David Chung Lin
Feb 2 at 3:36
$begingroup$
oh okay. Thank you. I will give that a try. I appreciate your time
$endgroup$
– MathIsHard
Feb 2 at 4:03
add a comment |
$begingroup$
Prove the central limit theorem for a sequence of i.i.d.
Bernoulli($p$) random variables, where $pin(0,1)$.
I am trying to do this by computing the moment generating function of the object I want the limit of and use Taylor's expansion to show that it converges to the moment generating function of a standard normal.
Attempt:
For a random variable X, we have the moment generating function $$M_{X}(t)=mathbb{E}[e^{tX}]$$ and if we expand using the Taylor series of $e^{tX}$ we get $$M_X(t)=sum_{n=0}^infty frac{mathbb{E}[X^n]}{n!}t^n.$$ So, we have in particular $$M_x^{(n)}(0)=mathbb{E}[X^n].$$
Now for the proof, we have a Bernoulli random variable is 1 with probability $p$ and 0 with probability $(1-p)$. First we want the moment generating function for a Bernoulli random variable. In particular, we have $$M_{Bernoulli(p)}=(1-p)+pe^t=1+(e^t-1)p$$
Then, we have a binomial random variable as the sum of n independent Bernoulli variables. So,
$$M_{binomial(n,p)}=((1-p)+pe^t=(1+(e^t-1)p)^n$$
Suppose that $p=lambda/n$ and notice that $$M_{binomial(n,lambda/n)}=bigg(1+frac{(e^t-1)lambda}{n}bigg)^n rightarrow e^{lambda(e^t-1)}$$
I am not sure how to finish the proof though or if I am even on the right track...
probability probability-theory random-variables central-limit-theorem
asked Feb 2 at 1:18
MathIsHardMathIsHard
1,318516
$endgroup$
Prove the central limit theorem for a sequence of i.i.d.
Bernoulli($p$) random variables, where $pin(0,1)$.
I am trying to do this by computing the moment generating function of the object I want the limit of and use Taylor's expansion to show that it converges to the moment generating function of a standard normal.
Attempt:
For a random variable X, we have the moment generating function $$M_{X}(t)=mathbb{E}[e^{tX}]$$ and if we expand using the Taylor series of $e^{tX}$ we get $$M_X(t)=sum_{n=0}^infty frac{mathbb{E}[X^n]}{n!}t^n.$$ So, we have in particular $$M_x^{(n)}(0)=mathbb{E}[X^n].$$
Now for the proof, we have a Bernoulli random variable is 1 with probability $p$ and 0 with probability $(1-p)$. First we want the moment generating function for a Bernoulli random variable. In particular, we have $$M_{Bernoulli(p)}=(1-p)+pe^t=1+(e^t-1)p$$
Then, we have a binomial random variable as the sum of n independent Bernoulli variables. So,
$$M_{binomial(n,p)}=((1-p)+pe^t=(1+(e^t-1)p)^n$$
Suppose that $p=lambda/n$ and notice that $$M_{binomial(n,lambda/n)}=bigg(1+frac{(e^t-1)lambda}{n}bigg)^n rightarrow e^{lambda(e^t-1)}$$
I am not sure how to finish the proof though or if I am even on the right track...
probability probability-theory random-variables central-limit-theorem
probability probability-theory random-variables central-limit-theorem
asked Feb 2 at 1:18
MathIsHardMathIsHard
1,318516
asked Feb 2 at 1:18
MathIsHardMathIsHard
1,318516
asked Feb 2 at 1:18
MathIsHardMathIsHard
1,318516
asked Feb 2 at 1:18
MathIsHardMathIsHard
1,318516
asked Feb 2 at 1:18
MathIsHardMathIsHard
1,318516
1,318516
1
$begingroup$
What you have done so far is merely deriving (correctly) Binomial becoming Poisson under the limit $n to infty, p to 0$ with $np = lambda$ held constant. The statement of Central Limit Theorem involves $frac{bar X - mu}{ sigma}$. Where's your analysis on $bar X$?
$endgroup$
– Lee David Chung Lin
Feb 2 at 2:05
$begingroup$
Oh okay. Thank you. So I want to find the mean and the std deviation for the random variable next?
$endgroup$
– MathIsHard
Feb 2 at 2:48
1
$begingroup$
There are many ways to do this. Since you wanted to use the moment generating function, you should find the MGF of $M_Y(t)$ where $Y equiv bar X = frac1n sum X_i$ and then further consider the MGF of $sqrt{n}frac{Y - mu}{ sigma}$
$endgroup$
– Lee David Chung Lin
Feb 2 at 3:36
$begingroup$
oh okay. Thank you. I will give that a try. I appreciate your time
$endgroup$
– MathIsHard
Feb 2 at 4:03
add a comment |
1
$begingroup$
What you have done so far is merely deriving (correctly) Binomial becoming Poisson under the limit $n to infty, p to 0$ with $np = lambda$ held constant. The statement of Central Limit Theorem involves $frac{bar X - mu}{ sigma}$. Where's your analysis on $bar X$?
$endgroup$
– Lee David Chung Lin
Feb 2 at 2:05
$begingroup$
Oh okay. Thank you. So I want to find the mean and the std deviation for the random variable next?
$endgroup$
– MathIsHard
Feb 2 at 2:48
1
$begingroup$
There are many ways to do this. Since you wanted to use the moment generating function, you should find the MGF of $M_Y(t)$ where $Y equiv bar X = frac1n sum X_i$ and then further consider the MGF of $sqrt{n}frac{Y - mu}{ sigma}$
$endgroup$
– Lee David Chung Lin
Feb 2 at 3:36
$begingroup$
oh okay. Thank you. I will give that a try. I appreciate your time
$endgroup$
– MathIsHard
Feb 2 at 4:03
1
1
$begingroup$
What you have done so far is merely deriving (correctly) Binomial becoming Poisson under the limit $n to infty, p to 0$ with $np = lambda$ held constant. The statement of Central Limit Theorem involves $frac{bar X - mu}{ sigma}$. Where's your analysis on $bar X$?
$endgroup$
– Lee David Chung Lin
Feb 2 at 2:05
$begingroup$
What you have done so far is merely deriving (correctly) Binomial becoming Poisson under the limit $n to infty, p to 0$ with $np = lambda$ held constant. The statement of Central Limit Theorem involves $frac{bar X - mu}{ sigma}$. Where's your analysis on $bar X$?
$endgroup$
– Lee David Chung Lin
Feb 2 at 2:05
$begingroup$
Oh okay. Thank you. So I want to find the mean and the std deviation for the random variable next?
$endgroup$
– MathIsHard
Feb 2 at 2:48
$begingroup$
Oh okay. Thank you. So I want to find the mean and the std deviation for the random variable next?
$endgroup$
– MathIsHard
Feb 2 at 2:48
1
1
$begingroup$
There are many ways to do this. Since you wanted to use the moment generating function, you should find the MGF of $M_Y(t)$ where $Y equiv bar X = frac1n sum X_i$ and then further consider the MGF of $sqrt{n}frac{Y - mu}{ sigma}$
$endgroup$
– Lee David Chung Lin
Feb 2 at 3:36
$begingroup$
There are many ways to do this. Since you wanted to use the moment generating function, you should find the MGF of $M_Y(t)$ where $Y equiv bar X = frac1n sum X_i$ and then further consider the MGF of $sqrt{n}frac{Y - mu}{ sigma}$
$endgroup$
– Lee David Chung Lin
Feb 2 at 3:36
$begingroup$
oh okay. Thank you. I will give that a try. I appreciate your time
$endgroup$
– MathIsHard
Feb 2 at 4:03
$begingroup$
oh okay. Thank you. I will give that a try. I appreciate your time
$endgroup$
– MathIsHard
Feb 2 at 4:03
add a comment |
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$begingroup$
What you have done so far is merely deriving (correctly) Binomial becoming Poisson under the limit $n to infty, p to 0$ with $np = lambda$ held constant. The statement of Central Limit Theorem involves $frac{bar X - mu}{ sigma}$. Where's your analysis on $bar X$?
$endgroup$
– Lee David Chung Lin
Feb 2 at 2:05
$begingroup$
Oh okay. Thank you. So I want to find the mean and the std deviation for the random variable next?
$endgroup$
– MathIsHard
Feb 2 at 2:48
1
$begingroup$
There are many ways to do this. Since you wanted to use the moment generating function, you should find the MGF of $M_Y(t)$ where $Y equiv bar X = frac1n sum X_i$ and then further consider the MGF of $sqrt{n}frac{Y - mu}{ sigma}$
$endgroup$
– Lee David Chung Lin
Feb 2 at 3:36
$begingroup$
oh okay. Thank you. I will give that a try. I appreciate your time
$endgroup$
– MathIsHard
Feb 2 at 4:03