Distribution of $bar{X}$ of n Bernoulli's
$begingroup$
I am trying to derive the distribution of $bar{X}_n$ where $X_1, X_2,...,X_n$ are iid $sim mathrm{Bern}(p)$. I used two approaches but I am debating myself and questioning which one would be correct (if any).
Method 1: Using MGF
I used the moment generating function and ended up with $bar{X}_n sim mathrm{Bern}(p^n)$
Method 2:
I used the CLT and ended up with $bar{X}_n sim N(p, sqrt{pq}/n$) for n being large.
I am not sure which one is correct (if any).
Can someone tell me if I am doing this right or not?
Thank you,
I appreciate your help
probability-theory probability-distributions
edited Dec 5 '18 at 5:29
epimorphic
2,74131533
asked Dec 4 '18 at 0:56
S AS A
1
$endgroup$
add a comment |
$begingroup$
I am trying to derive the distribution of $bar{X}_n$ where $X_1, X_2,...,X_n$ are iid $sim mathrm{Bern}(p)$. I used two approaches but I am debating myself and questioning which one would be correct (if any).
Method 1: Using MGF
I used the moment generating function and ended up with $bar{X}_n sim mathrm{Bern}(p^n)$
Method 2:
I used the CLT and ended up with $bar{X}_n sim N(p, sqrt{pq}/n$) for n being large.
I am not sure which one is correct (if any).
Can someone tell me if I am doing this right or not?
Thank you,
I appreciate your help
probability-theory probability-distributions
edited Dec 5 '18 at 5:29
epimorphic
2,74131533
asked Dec 4 '18 at 0:56
S AS A
1
$endgroup$
add a comment |
$begingroup$
I am trying to derive the distribution of $bar{X}_n$ where $X_1, X_2,...,X_n$ are iid $sim mathrm{Bern}(p)$. I used two approaches but I am debating myself and questioning which one would be correct (if any).
Method 1: Using MGF
I used the moment generating function and ended up with $bar{X}_n sim mathrm{Bern}(p^n)$
Method 2:
I used the CLT and ended up with $bar{X}_n sim N(p, sqrt{pq}/n$) for n being large.
I am not sure which one is correct (if any).
Can someone tell me if I am doing this right or not?
Thank you,
I appreciate your help
probability-theory probability-distributions
edited Dec 5 '18 at 5:29
epimorphic
2,74131533
asked Dec 4 '18 at 0:56
S AS A
1
$endgroup$
I am trying to derive the distribution of $bar{X}_n$ where $X_1, X_2,...,X_n$ are iid $sim mathrm{Bern}(p)$. I used two approaches but I am debating myself and questioning which one would be correct (if any).
Method 1: Using MGF
I used the moment generating function and ended up with $bar{X}_n sim mathrm{Bern}(p^n)$
Method 2:
I used the CLT and ended up with $bar{X}_n sim N(p, sqrt{pq}/n$) for n being large.
I am not sure which one is correct (if any).
Can someone tell me if I am doing this right or not?
Thank you,
I appreciate your help
probability-theory probability-distributions
probability-theory probability-distributions
edited Dec 5 '18 at 5:29
epimorphic
2,74131533
asked Dec 4 '18 at 0:56
S AS A
1
edited Dec 5 '18 at 5:29
epimorphic
2,74131533
asked Dec 4 '18 at 0:56
S AS A
1
edited Dec 5 '18 at 5:29
epimorphic
2,74131533
edited Dec 5 '18 at 5:29
epimorphic
2,74131533
edited Dec 5 '18 at 5:29
epimorphic
2,74131533
2,74131533
asked Dec 4 '18 at 0:56
S AS A
1
asked Dec 4 '18 at 0:56
S AS A
1
asked Dec 4 '18 at 0:56
S AS A
1
1
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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Neither.
For the first one, you are claiming that the average only take binary value.
For the second one, what if $n$ is small.
Guide:
- Recall that sum of IID Bernoulli follows a binomial distribution $Bin(n, p)$.
- Average is simply dividing the sum by $n$.
answered Dec 4 '18 at 1:04
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
$endgroup$
– S A
Dec 4 '18 at 1:10
$begingroup$
Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
$endgroup$
– Siong Thye Goh
Dec 4 '18 at 1:12
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
Neither.
For the first one, you are claiming that the average only take binary value.
For the second one, what if $n$ is small.
Guide:
- Recall that sum of IID Bernoulli follows a binomial distribution $Bin(n, p)$.
- Average is simply dividing the sum by $n$.
answered Dec 4 '18 at 1:04
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
$endgroup$
– S A
Dec 4 '18 at 1:10
$begingroup$
Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
$endgroup$
– Siong Thye Goh
Dec 4 '18 at 1:12
add a comment |
$begingroup$
Neither.
For the first one, you are claiming that the average only take binary value.
For the second one, what if $n$ is small.
Guide:
- Recall that sum of IID Bernoulli follows a binomial distribution $Bin(n, p)$.
- Average is simply dividing the sum by $n$.
answered Dec 4 '18 at 1:04
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
$endgroup$
– S A
Dec 4 '18 at 1:10
$begingroup$
Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
$endgroup$
– Siong Thye Goh
Dec 4 '18 at 1:12
add a comment |
$begingroup$
Neither.
For the first one, you are claiming that the average only take binary value.
For the second one, what if $n$ is small.
Guide:
- Recall that sum of IID Bernoulli follows a binomial distribution $Bin(n, p)$.
- Average is simply dividing the sum by $n$.
answered Dec 4 '18 at 1:04
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
Neither.
For the first one, you are claiming that the average only take binary value.
For the second one, what if $n$ is small.
Guide:
- Recall that sum of IID Bernoulli follows a binomial distribution $Bin(n, p)$.
- Average is simply dividing the sum by $n$.
answered Dec 4 '18 at 1:04
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 4 '18 at 1:04
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 4 '18 at 1:04
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 4 '18 at 1:04
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
$begingroup$
Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
$endgroup$
– S A
Dec 4 '18 at 1:10
$begingroup$
Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
$endgroup$
– Siong Thye Goh
Dec 4 '18 at 1:12
add a comment |
$begingroup$
Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
$endgroup$
– S A
Dec 4 '18 at 1:10
$begingroup$
Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
$endgroup$
– Siong Thye Goh
Dec 4 '18 at 1:12
$begingroup$
Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
$endgroup$
– S A
Dec 4 '18 at 1:10
$begingroup$
Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
$endgroup$
– S A
Dec 4 '18 at 1:10
$begingroup$
Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
$endgroup$
– Siong Thye Goh
Dec 4 '18 at 1:12
$begingroup$
Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
$endgroup$
– Siong Thye Goh
Dec 4 '18 at 1:12
add a comment |
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