Mixing
Proof involving expected values
$begingroup$
Let $X_1, ..., X_n$ be IID random variables with mean $mu$, standard deviation $sigma$ and finite fourth moment. Prove by induction that
$E([sum_{i=1}^{n} (x_i - mu)]^{4}) = nE((x_i-mu)^{4})+6{nchoose 2}sigma^{4}$.
First, I can rewrite this equation as follows (since they are IID, each $x_i$ is the same distribution as any other):
$E([sum_{i=1}^{n} (x_i - mu)]^{4}) = nE((x-mu)^{4})+6{nchoose 2}sigma^{4}$.
Note the small change in the first part of the RHS.
Base case: $n=1$
$E((x - mu)^{4}) = (1)E((x-mu)^{4})+6{1choose 2}sigma^{4}$. But ${n choose k} = 0$ when $k > n$, so we get the desired result.
Induction Hypothesis: Suppose this result holds for $n=1, ..., k$. We show it holds for $k+1$. Now I plug in $n=k+1$, as follows:
begin{align*}
E([sum_{i=1}^{k+1} (x_i-mu)]^{4}) = E([sum_{i=1}^{k} (x_i-mu) + (x-mu)]^{4}).
end{align*}
Using the binomial theorem with $a = sum_{i=1}^{k} (x_i-mu)$ and $b =
(x-mu)$, as well as the fact that $E[X+Y] = E[X]+E[Y]$, and $E[XY] = E[X]E[Y]$ (from independence), I have:
begin{align*}
E([sum_{i=1}^{k} x_i-mu]^{4}) + 4E([sum_{i=1}^{k} x_i-mu]^{3})E(x-mu)+6E([sum_{i=1}^{k} x_i-mu]^{2})E([x-mu]^{2})+4E([sum_{i=1}^{k} x_i-mu])E([x-mu]^{3})+E([x-mu]^{4}).
end{align*}
From the induction hypothesis I can say that the first term + the last term is equal to
begin{align*}
kE([x-mu]^{4})+6{kchoose 2}sigma^{4} + E([x-mu]^{4}) = (k+1)E([x-mu]^{4})+6{k choose 2}sigma^{4}.
end{align*}
This looks similar to what I want, I just need a way to obtain $6{k+1choose 2}sigma^{4}$.
$textbf{Edit}$: I think that $E(x-mu) = 0$, so the second term in the binomial expansion vanishes. So I just need to deal with the terms $6E([sum_{i=1}^{k} x_i-mu]^{2})E([x-mu]^{2})+4E([sum_{i=1}^{k} x_i-mu])E([x-mu]^{3})$.
probability-theory
asked Jan 22 at 19:44
mXdXmXdX
848
$endgroup$
|
show 1 more comment
$begingroup$
Let $X_1, ..., X_n$ be IID random variables with mean $mu$, standard deviation $sigma$ and finite fourth moment. Prove by induction that
$E([sum_{i=1}^{n} (x_i - mu)]^{4}) = nE((x_i-mu)^{4})+6{nchoose 2}sigma^{4}$.
First, I can rewrite this equation as follows (since they are IID, each $x_i$ is the same distribution as any other):
$E([sum_{i=1}^{n} (x_i - mu)]^{4}) = nE((x-mu)^{4})+6{nchoose 2}sigma^{4}$.
Note the small change in the first part of the RHS.
Base case: $n=1$
$E((x - mu)^{4}) = (1)E((x-mu)^{4})+6{1choose 2}sigma^{4}$. But ${n choose k} = 0$ when $k > n$, so we get the desired result.
Induction Hypothesis: Suppose this result holds for $n=1, ..., k$. We show it holds for $k+1$. Now I plug in $n=k+1$, as follows:
begin{align*}
E([sum_{i=1}^{k+1} (x_i-mu)]^{4}) = E([sum_{i=1}^{k} (x_i-mu) + (x-mu)]^{4}).
end{align*}
Using the binomial theorem with $a = sum_{i=1}^{k} (x_i-mu)$ and $b =
(x-mu)$, as well as the fact that $E[X+Y] = E[X]+E[Y]$, and $E[XY] = E[X]E[Y]$ (from independence), I have:
begin{align*}
E([sum_{i=1}^{k} x_i-mu]^{4}) + 4E([sum_{i=1}^{k} x_i-mu]^{3})E(x-mu)+6E([sum_{i=1}^{k} x_i-mu]^{2})E([x-mu]^{2})+4E([sum_{i=1}^{k} x_i-mu])E([x-mu]^{3})+E([x-mu]^{4}).
end{align*}
From the induction hypothesis I can say that the first term + the last term is equal to
begin{align*}
kE([x-mu]^{4})+6{kchoose 2}sigma^{4} + E([x-mu]^{4}) = (k+1)E([x-mu]^{4})+6{k choose 2}sigma^{4}.
end{align*}
This looks similar to what I want, I just need a way to obtain $6{k+1choose 2}sigma^{4}$.
$textbf{Edit}$: I think that $E(x-mu) = 0$, so the second term in the binomial expansion vanishes. So I just need to deal with the terms $6E([sum_{i=1}^{k} x_i-mu]^{2})E([x-mu]^{2})+4E([sum_{i=1}^{k} x_i-mu])E([x-mu]^{3})$.
probability-theory
asked Jan 22 at 19:44
mXdXmXdX
848
$endgroup$
$begingroup$
Necessarily by induction? 'Cause the direct proof is... well, direct.
$endgroup$
– Did
Jan 22 at 19:46
$begingroup$
Yes, I'm self-studying and the problem asks by induction. I thought it was direct as well, but had trouble doing it by induction.
$endgroup$
– mXdX
Jan 22 at 19:47
$begingroup$
I may be a bit slow... But I don’t even understand your first equation. With $i$ as a sum index on the LHS, $i$ alone on the RHS. And do you have hypothesis on the $x_i$?
$endgroup$
– mathcounterexamples.net
Jan 22 at 19:52
$begingroup$
Sorry, I've edited it. I left important information out. The $x_i$'s are IID, so the single $x_i$ can be re-written as $x$, for example.
$endgroup$
– mXdX
Jan 22 at 20:19
3
$begingroup$
$(a+b)^4=a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4$ is a good start
$endgroup$
– P. Quinton
Jan 22 at 20:49
|
show 1 more comment
$begingroup$
Let $X_1, ..., X_n$ be IID random variables with mean $mu$, standard deviation $sigma$ and finite fourth moment. Prove by induction that
$E([sum_{i=1}^{n} (x_i - mu)]^{4}) = nE((x_i-mu)^{4})+6{nchoose 2}sigma^{4}$.
First, I can rewrite this equation as follows (since they are IID, each $x_i$ is the same distribution as any other):
$E([sum_{i=1}^{n} (x_i - mu)]^{4}) = nE((x-mu)^{4})+6{nchoose 2}sigma^{4}$.
Note the small change in the first part of the RHS.
Base case: $n=1$
$E((x - mu)^{4}) = (1)E((x-mu)^{4})+6{1choose 2}sigma^{4}$. But ${n choose k} = 0$ when $k > n$, so we get the desired result.
Induction Hypothesis: Suppose this result holds for $n=1, ..., k$. We show it holds for $k+1$. Now I plug in $n=k+1$, as follows:
begin{align*}
E([sum_{i=1}^{k+1} (x_i-mu)]^{4}) = E([sum_{i=1}^{k} (x_i-mu) + (x-mu)]^{4}).
end{align*}
Using the binomial theorem with $a = sum_{i=1}^{k} (x_i-mu)$ and $b =
(x-mu)$, as well as the fact that $E[X+Y] = E[X]+E[Y]$, and $E[XY] = E[X]E[Y]$ (from independence), I have:
begin{align*}
E([sum_{i=1}^{k} x_i-mu]^{4}) + 4E([sum_{i=1}^{k} x_i-mu]^{3})E(x-mu)+6E([sum_{i=1}^{k} x_i-mu]^{2})E([x-mu]^{2})+4E([sum_{i=1}^{k} x_i-mu])E([x-mu]^{3})+E([x-mu]^{4}).
end{align*}
From the induction hypothesis I can say that the first term + the last term is equal to
begin{align*}
kE([x-mu]^{4})+6{kchoose 2}sigma^{4} + E([x-mu]^{4}) = (k+1)E([x-mu]^{4})+6{k choose 2}sigma^{4}.
end{align*}
This looks similar to what I want, I just need a way to obtain $6{k+1choose 2}sigma^{4}$.
$textbf{Edit}$: I think that $E(x-mu) = 0$, so the second term in the binomial expansion vanishes. So I just need to deal with the terms $6E([sum_{i=1}^{k} x_i-mu]^{2})E([x-mu]^{2})+4E([sum_{i=1}^{k} x_i-mu])E([x-mu]^{3})$.
probability-theory
asked Jan 22 at 19:44
mXdXmXdX
848
$endgroup$
Let $X_1, ..., X_n$ be IID random variables with mean $mu$, standard deviation $sigma$ and finite fourth moment. Prove by induction that
$E([sum_{i=1}^{n} (x_i - mu)]^{4}) = nE((x_i-mu)^{4})+6{nchoose 2}sigma^{4}$.
First, I can rewrite this equation as follows (since they are IID, each $x_i$ is the same distribution as any other):
$E([sum_{i=1}^{n} (x_i - mu)]^{4}) = nE((x-mu)^{4})+6{nchoose 2}sigma^{4}$.
Note the small change in the first part of the RHS.
Base case: $n=1$
$E((x - mu)^{4}) = (1)E((x-mu)^{4})+6{1choose 2}sigma^{4}$. But ${n choose k} = 0$ when $k > n$, so we get the desired result.
Induction Hypothesis: Suppose this result holds for $n=1, ..., k$. We show it holds for $k+1$. Now I plug in $n=k+1$, as follows:
begin{align*}
E([sum_{i=1}^{k+1} (x_i-mu)]^{4}) = E([sum_{i=1}^{k} (x_i-mu) + (x-mu)]^{4}).
end{align*}
Using the binomial theorem with $a = sum_{i=1}^{k} (x_i-mu)$ and $b =
(x-mu)$, as well as the fact that $E[X+Y] = E[X]+E[Y]$, and $E[XY] = E[X]E[Y]$ (from independence), I have:
begin{align*}
E([sum_{i=1}^{k} x_i-mu]^{4}) + 4E([sum_{i=1}^{k} x_i-mu]^{3})E(x-mu)+6E([sum_{i=1}^{k} x_i-mu]^{2})E([x-mu]^{2})+4E([sum_{i=1}^{k} x_i-mu])E([x-mu]^{3})+E([x-mu]^{4}).
end{align*}
From the induction hypothesis I can say that the first term + the last term is equal to
begin{align*}
kE([x-mu]^{4})+6{kchoose 2}sigma^{4} + E([x-mu]^{4}) = (k+1)E([x-mu]^{4})+6{k choose 2}sigma^{4}.
end{align*}
This looks similar to what I want, I just need a way to obtain $6{k+1choose 2}sigma^{4}$.
$textbf{Edit}$: I think that $E(x-mu) = 0$, so the second term in the binomial expansion vanishes. So I just need to deal with the terms $6E([sum_{i=1}^{k} x_i-mu]^{2})E([x-mu]^{2})+4E([sum_{i=1}^{k} x_i-mu])E([x-mu]^{3})$.
probability-theory
probability-theory
asked Jan 22 at 19:44
mXdXmXdX
848
asked Jan 22 at 19:44
mXdXmXdX
848
edited Jan 23 at 6:52
mXdX
asked Jan 22 at 19:44
mXdXmXdX
848
asked Jan 22 at 19:44
mXdXmXdX
848
asked Jan 22 at 19:44
mXdXmXdX
848
848
$begingroup$
Necessarily by induction? 'Cause the direct proof is... well, direct.
$endgroup$
– Did
Jan 22 at 19:46
$begingroup$
Yes, I'm self-studying and the problem asks by induction. I thought it was direct as well, but had trouble doing it by induction.
$endgroup$
– mXdX
Jan 22 at 19:47
$begingroup$
I may be a bit slow... But I don’t even understand your first equation. With $i$ as a sum index on the LHS, $i$ alone on the RHS. And do you have hypothesis on the $x_i$?
$endgroup$
– mathcounterexamples.net
Jan 22 at 19:52
$begingroup$
Sorry, I've edited it. I left important information out. The $x_i$'s are IID, so the single $x_i$ can be re-written as $x$, for example.
$endgroup$
– mXdX
Jan 22 at 20:19
3
$begingroup$
$(a+b)^4=a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4$ is a good start
$endgroup$
– P. Quinton
Jan 22 at 20:49
|
show 1 more comment
$begingroup$
Necessarily by induction? 'Cause the direct proof is... well, direct.
$endgroup$
– Did
Jan 22 at 19:46
$begingroup$
Yes, I'm self-studying and the problem asks by induction. I thought it was direct as well, but had trouble doing it by induction.
$endgroup$
– mXdX
Jan 22 at 19:47
$begingroup$
I may be a bit slow... But I don’t even understand your first equation. With $i$ as a sum index on the LHS, $i$ alone on the RHS. And do you have hypothesis on the $x_i$?
$endgroup$
– mathcounterexamples.net
Jan 22 at 19:52
$begingroup$
Sorry, I've edited it. I left important information out. The $x_i$'s are IID, so the single $x_i$ can be re-written as $x$, for example.
$endgroup$
– mXdX
Jan 22 at 20:19
3
$begingroup$
$(a+b)^4=a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4$ is a good start
$endgroup$
– P. Quinton
Jan 22 at 20:49
$begingroup$
Necessarily by induction? 'Cause the direct proof is... well, direct.
$endgroup$
– Did
Jan 22 at 19:46
$begingroup$
Necessarily by induction? 'Cause the direct proof is... well, direct.
$endgroup$
– Did
Jan 22 at 19:46
$begingroup$
Yes, I'm self-studying and the problem asks by induction. I thought it was direct as well, but had trouble doing it by induction.
$endgroup$
– mXdX
Jan 22 at 19:47
$begingroup$
Yes, I'm self-studying and the problem asks by induction. I thought it was direct as well, but had trouble doing it by induction.
$endgroup$
– mXdX
Jan 22 at 19:47
$begingroup$
I may be a bit slow... But I don’t even understand your first equation. With $i$ as a sum index on the LHS, $i$ alone on the RHS. And do you have hypothesis on the $x_i$?
$endgroup$
– mathcounterexamples.net
Jan 22 at 19:52
$begingroup$
I may be a bit slow... But I don’t even understand your first equation. With $i$ as a sum index on the LHS, $i$ alone on the RHS. And do you have hypothesis on the $x_i$?
$endgroup$
– mathcounterexamples.net
Jan 22 at 19:52
$begingroup$
Sorry, I've edited it. I left important information out. The $x_i$'s are IID, so the single $x_i$ can be re-written as $x$, for example.
$endgroup$
– mXdX
Jan 22 at 20:19
$begingroup$
Sorry, I've edited it. I left important information out. The $x_i$'s are IID, so the single $x_i$ can be re-written as $x$, for example.
$endgroup$
– mXdX
Jan 22 at 20:19
3
3
$begingroup$
$(a+b)^4=a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4$ is a good start
$endgroup$
– P. Quinton
Jan 22 at 20:49
$begingroup$
$(a+b)^4=a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4$ is a good start
$endgroup$
– P. Quinton
Jan 22 at 20:49
|
show 1 more comment
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$begingroup$
Necessarily by induction? 'Cause the direct proof is... well, direct.
$endgroup$
– Did
Jan 22 at 19:46
$begingroup$
Yes, I'm self-studying and the problem asks by induction. I thought it was direct as well, but had trouble doing it by induction.
$endgroup$
– mXdX
Jan 22 at 19:47
$begingroup$
I may be a bit slow... But I don’t even understand your first equation. With $i$ as a sum index on the LHS, $i$ alone on the RHS. And do you have hypothesis on the $x_i$?
$endgroup$
– mathcounterexamples.net
Jan 22 at 19:52
$begingroup$
Sorry, I've edited it. I left important information out. The $x_i$'s are IID, so the single $x_i$ can be re-written as $x$, for example.
$endgroup$
– mXdX
Jan 22 at 20:19
3
$begingroup$
$(a+b)^4=a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4$ is a good start
$endgroup$
– P. Quinton
Jan 22 at 20:49