Variance of Binomial Distribution $E[X^2]$
up vote
4
down vote
favorite
so I'm trying to use the equation:
$Var(X) = E[X^2] - (E[x])^2$,
And for the $E[X^2]$ part, I'm trying to use the method of indicators...
However, when I do that, I get the same value as with $E[X]$...
Is it wrong to try to use the method of indicators for this case?
Basically, I end up with
$Var(X) = np(1-np)$,
when it should be $Var(X) = np(1-p)$.
probability
edited Mar 6 '13 at 5:49
Amzoti
50.8k125397
asked Mar 6 '13 at 5:48
Joy
23113
add a comment |
up vote
4
down vote
favorite
so I'm trying to use the equation:
$Var(X) = E[X^2] - (E[x])^2$,
And for the $E[X^2]$ part, I'm trying to use the method of indicators...
However, when I do that, I get the same value as with $E[X]$...
Is it wrong to try to use the method of indicators for this case?
Basically, I end up with
$Var(X) = np(1-np)$,
when it should be $Var(X) = np(1-p)$.
probability
edited Mar 6 '13 at 5:49
Amzoti
50.8k125397
asked Mar 6 '13 at 5:48
Joy
23113
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
so I'm trying to use the equation:
$Var(X) = E[X^2] - (E[x])^2$,
And for the $E[X^2]$ part, I'm trying to use the method of indicators...
However, when I do that, I get the same value as with $E[X]$...
Is it wrong to try to use the method of indicators for this case?
Basically, I end up with
$Var(X) = np(1-np)$,
when it should be $Var(X) = np(1-p)$.
probability
edited Mar 6 '13 at 5:49
Amzoti
50.8k125397
asked Mar 6 '13 at 5:48
Joy
23113
so I'm trying to use the equation:
$Var(X) = E[X^2] - (E[x])^2$,
And for the $E[X^2]$ part, I'm trying to use the method of indicators...
However, when I do that, I get the same value as with $E[X]$...
Is it wrong to try to use the method of indicators for this case?
Basically, I end up with
$Var(X) = np(1-np)$,
when it should be $Var(X) = np(1-p)$.
probability
probability
edited Mar 6 '13 at 5:49
Amzoti
50.8k125397
asked Mar 6 '13 at 5:48
Joy
23113
edited Mar 6 '13 at 5:49
Amzoti
50.8k125397
asked Mar 6 '13 at 5:48
Joy
23113
edited Mar 6 '13 at 5:49
Amzoti
50.8k125397
edited Mar 6 '13 at 5:49
Amzoti
50.8k125397
edited Mar 6 '13 at 5:49
Amzoti
50.8k125397
50.8k125397
asked Mar 6 '13 at 5:48
Joy
23113
asked Mar 6 '13 at 5:48
Joy
23113
asked Mar 6 '13 at 5:48
Joy
23113
23113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
9
down vote
accepted
The method of indicators works well here. Let $X_i=1$ if we have a success on the $i$-th trial, and $0$ otherwise.
Then $X=X_1+X_2+cdots+X_n$. Expand $(X_1+X_2+cdots+X_n)^2$. We find that
$$X^2=sum_1^n X_i^2+2sum_{ilt j}X_iX_j.$$
Calculate the expectation, using linearity. Each $X_i^2$ has expectation $p$, since $X_i^2=X_i$. Thus $E(sum_1^n X_i^2)=np$.
By independence, if $ine j$, $E(X_i X_j)=E(X_i)E(X_j)=p^2$.
Now count. The number of pairs $(i,j)$ with $ilt j$ is $binom{n}{2}$. So we get
$$E(X^2)=np +n(n-1)p^2.$$
But $(E(X))^2=n^2p^2$. Thus $E(X^2)-(E(X))^2=np-np^2$. This simplifies to $np(1-p)$.
answered Mar 6 '13 at 6:05
André Nicolas
450k36421805
You are welcome. The evaluation of $E(X_iX_j)$ comes up fairly often when we use the method of indicator functions. In this case, things were simple, because of independence. However, there are plenty of situations (a simple example is the hypergeometric) where independence fails, but $E(X_iX_j)$ is accessible.
– André Nicolas
Mar 6 '13 at 6:57
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
The method of indicators works well here. Let $X_i=1$ if we have a success on the $i$-th trial, and $0$ otherwise.
Then $X=X_1+X_2+cdots+X_n$. Expand $(X_1+X_2+cdots+X_n)^2$. We find that
$$X^2=sum_1^n X_i^2+2sum_{ilt j}X_iX_j.$$
Calculate the expectation, using linearity. Each $X_i^2$ has expectation $p$, since $X_i^2=X_i$. Thus $E(sum_1^n X_i^2)=np$.
By independence, if $ine j$, $E(X_i X_j)=E(X_i)E(X_j)=p^2$.
Now count. The number of pairs $(i,j)$ with $ilt j$ is $binom{n}{2}$. So we get
$$E(X^2)=np +n(n-1)p^2.$$
But $(E(X))^2=n^2p^2$. Thus $E(X^2)-(E(X))^2=np-np^2$. This simplifies to $np(1-p)$.
answered Mar 6 '13 at 6:05
André Nicolas
450k36421805
You are welcome. The evaluation of $E(X_iX_j)$ comes up fairly often when we use the method of indicator functions. In this case, things were simple, because of independence. However, there are plenty of situations (a simple example is the hypergeometric) where independence fails, but $E(X_iX_j)$ is accessible.
– André Nicolas
Mar 6 '13 at 6:57
add a comment |
up vote
9
down vote
accepted
The method of indicators works well here. Let $X_i=1$ if we have a success on the $i$-th trial, and $0$ otherwise.
Then $X=X_1+X_2+cdots+X_n$. Expand $(X_1+X_2+cdots+X_n)^2$. We find that
$$X^2=sum_1^n X_i^2+2sum_{ilt j}X_iX_j.$$
Calculate the expectation, using linearity. Each $X_i^2$ has expectation $p$, since $X_i^2=X_i$. Thus $E(sum_1^n X_i^2)=np$.
By independence, if $ine j$, $E(X_i X_j)=E(X_i)E(X_j)=p^2$.
Now count. The number of pairs $(i,j)$ with $ilt j$ is $binom{n}{2}$. So we get
$$E(X^2)=np +n(n-1)p^2.$$
But $(E(X))^2=n^2p^2$. Thus $E(X^2)-(E(X))^2=np-np^2$. This simplifies to $np(1-p)$.
answered Mar 6 '13 at 6:05
André Nicolas
450k36421805
You are welcome. The evaluation of $E(X_iX_j)$ comes up fairly often when we use the method of indicator functions. In this case, things were simple, because of independence. However, there are plenty of situations (a simple example is the hypergeometric) where independence fails, but $E(X_iX_j)$ is accessible.
– André Nicolas
Mar 6 '13 at 6:57
add a comment |
up vote
9
down vote
accepted
up vote
9
down vote
accepted
The method of indicators works well here. Let $X_i=1$ if we have a success on the $i$-th trial, and $0$ otherwise.
Then $X=X_1+X_2+cdots+X_n$. Expand $(X_1+X_2+cdots+X_n)^2$. We find that
$$X^2=sum_1^n X_i^2+2sum_{ilt j}X_iX_j.$$
Calculate the expectation, using linearity. Each $X_i^2$ has expectation $p$, since $X_i^2=X_i$. Thus $E(sum_1^n X_i^2)=np$.
By independence, if $ine j$, $E(X_i X_j)=E(X_i)E(X_j)=p^2$.
Now count. The number of pairs $(i,j)$ with $ilt j$ is $binom{n}{2}$. So we get
$$E(X^2)=np +n(n-1)p^2.$$
But $(E(X))^2=n^2p^2$. Thus $E(X^2)-(E(X))^2=np-np^2$. This simplifies to $np(1-p)$.
answered Mar 6 '13 at 6:05
André Nicolas
450k36421805
The method of indicators works well here. Let $X_i=1$ if we have a success on the $i$-th trial, and $0$ otherwise.
Then $X=X_1+X_2+cdots+X_n$. Expand $(X_1+X_2+cdots+X_n)^2$. We find that
$$X^2=sum_1^n X_i^2+2sum_{ilt j}X_iX_j.$$
Calculate the expectation, using linearity. Each $X_i^2$ has expectation $p$, since $X_i^2=X_i$. Thus $E(sum_1^n X_i^2)=np$.
By independence, if $ine j$, $E(X_i X_j)=E(X_i)E(X_j)=p^2$.
Now count. The number of pairs $(i,j)$ with $ilt j$ is $binom{n}{2}$. So we get
$$E(X^2)=np +n(n-1)p^2.$$
But $(E(X))^2=n^2p^2$. Thus $E(X^2)-(E(X))^2=np-np^2$. This simplifies to $np(1-p)$.
answered Mar 6 '13 at 6:05
André Nicolas
450k36421805
edited Mar 6 '13 at 6:12
answered Mar 6 '13 at 6:05
André Nicolas
450k36421805
answered Mar 6 '13 at 6:05
André Nicolas
450k36421805
answered Mar 6 '13 at 6:05
André Nicolas
450k36421805
450k36421805
You are welcome. The evaluation of $E(X_iX_j)$ comes up fairly often when we use the method of indicator functions. In this case, things were simple, because of independence. However, there are plenty of situations (a simple example is the hypergeometric) where independence fails, but $E(X_iX_j)$ is accessible.
– André Nicolas
Mar 6 '13 at 6:57
add a comment |
You are welcome. The evaluation of $E(X_iX_j)$ comes up fairly often when we use the method of indicator functions. In this case, things were simple, because of independence. However, there are plenty of situations (a simple example is the hypergeometric) where independence fails, but $E(X_iX_j)$ is accessible.
– André Nicolas
Mar 6 '13 at 6:57
You are welcome. The evaluation of $E(X_iX_j)$ comes up fairly often when we use the method of indicator functions. In this case, things were simple, because of independence. However, there are plenty of situations (a simple example is the hypergeometric) where independence fails, but $E(X_iX_j)$ is accessible.
– André Nicolas
Mar 6 '13 at 6:57
You are welcome. The evaluation of $E(X_iX_j)$ comes up fairly often when we use the method of indicator functions. In this case, things were simple, because of independence. However, there are plenty of situations (a simple example is the hypergeometric) where independence fails, but $E(X_iX_j)$ is accessible.
– André Nicolas
Mar 6 '13 at 6:57
add a comment |
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