Variance of a sum of IID random variables.

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Let's say we have a sequence of $n$ IID random variables, $I_i$. Let's define a new random variable which is their sum:

$$S = sum_i I_i$$

To calculate the variance of $S$ we can say -

$$V(S) = sum_i V(I_i) = nV(I_1)$$

Or we can also say that $S$ has the same distribution as $nI_1$

So, we should have $V(S) = V(nI_1) = n^2V(I_1)$.

This is of course, in direct contradiction to what we got above. What am I missing?

probability random-variables independence variance

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asked Jul 17 at 1:04

Rohit Pandey

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up vote
2
down vote

favorite

Let's say we have a sequence of $n$ IID random variables, $I_i$. Let's define a new random variable which is their sum:

$$S = sum_i I_i$$

To calculate the variance of $S$ we can say -

$$V(S) = sum_i V(I_i) = nV(I_1)$$

Or we can also say that $S$ has the same distribution as $nI_1$

So, we should have $V(S) = V(nI_1) = n^2V(I_1)$.

This is of course, in direct contradiction to what we got above. What am I missing?

probability random-variables independence variance

share|cite|improve this question

asked Jul 17 at 1:04

Rohit Pandey

798718

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let's say we have a sequence of $n$ IID random variables, $I_i$. Let's define a new random variable which is their sum:

$$S = sum_i I_i$$

To calculate the variance of $S$ we can say -

$$V(S) = sum_i V(I_i) = nV(I_1)$$

Or we can also say that $S$ has the same distribution as $nI_1$

So, we should have $V(S) = V(nI_1) = n^2V(I_1)$.

This is of course, in direct contradiction to what we got above. What am I missing?

probability random-variables independence variance

share|cite|improve this question

asked Jul 17 at 1:04

Rohit Pandey

798718

Let's say we have a sequence of $n$ IID random variables, $I_i$. Let's define a new random variable which is their sum:

$$S = sum_i I_i$$

To calculate the variance of $S$ we can say -

$$V(S) = sum_i V(I_i) = nV(I_1)$$

Or we can also say that $S$ has the same distribution as $nI_1$

So, we should have $V(S) = V(nI_1) = n^2V(I_1)$.

This is of course, in direct contradiction to what we got above. What am I missing?

probability random-variables independence variance

share|cite|improve this question

asked Jul 17 at 1:04

Rohit Pandey

798718

share|cite|improve this question

share|cite|improve this question

asked Jul 17 at 1:04

Rohit Pandey

798718

asked Jul 17 at 1:04

Rohit Pandey

798718

asked Jul 17 at 1:04

Rohit Pandey

798718

798718

add a comment | 

add a comment | 

3 Answers
3

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up vote
7
down vote



accepted

$S$ does not have the same distribution as $nI_1.$ In fact, you have proven this by showing they have different variances. As an example, if $I_1$ can take the values $0$ and $1$ (i.e. it is a Bernoulli variable), then $2I_1$ can take the values $0$ and $2.$ However $I_1+I_2$ can take the value $0$ (if both are zero), $1$ (if one is zero and the other is one), or $2$ (if both are one).

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answered Jul 17 at 1:19

spaceisdarkgreen

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up vote
6
down vote

You're missing the difference between these two things:
$$
beginalign
& I_1 + I_2+ cdots + I_n, \[8pt] & I_1 + I_1+ cdots + I_1.
endalign
$$
In one of those, the terms are independent; in the other, they're as far from independent as they can get.

NOTE: These do not both have the same distribution.

share|cite|improve this answer

answered Jul 17 at 3:51

Michael Hardy

204k23186462

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up vote
4
down vote

We can not say $S$ has the same distribution as $nI_1$.

Example:

Toss two coins and count the heads as $I_1,I_2$ and their sum as $S$.   Now $I_1in0,1$, as is $I_2$, and they are iid as required.   So $Sin 0,1,2$ and yet $2I_1in 0,2$.   Thus we see that $S$ and $2I_1$ do not have the same support, therefore they cannot have the same distribution.

share|cite|improve this answer

answered Jul 17 at 1:28

Graham Kemp

80.1k43275

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote



accepted

$S$ does not have the same distribution as $nI_1.$ In fact, you have proven this by showing they have different variances. As an example, if $I_1$ can take the values $0$ and $1$ (i.e. it is a Bernoulli variable), then $2I_1$ can take the values $0$ and $2.$ However $I_1+I_2$ can take the value $0$ (if both are zero), $1$ (if one is zero and the other is one), or $2$ (if both are one).

share|cite|improve this answer

answered Jul 17 at 1:19

spaceisdarkgreen

27.6k21547

add a comment | 

up vote
7
down vote



accepted

$S$ does not have the same distribution as $nI_1.$ In fact, you have proven this by showing they have different variances. As an example, if $I_1$ can take the values $0$ and $1$ (i.e. it is a Bernoulli variable), then $2I_1$ can take the values $0$ and $2.$ However $I_1+I_2$ can take the value $0$ (if both are zero), $1$ (if one is zero and the other is one), or $2$ (if both are one).

share|cite|improve this answer

answered Jul 17 at 1:19

spaceisdarkgreen

27.6k21547

add a comment | 

up vote
7
down vote



accepted

up vote
7
down vote



accepted

$S$ does not have the same distribution as $nI_1.$ In fact, you have proven this by showing they have different variances. As an example, if $I_1$ can take the values $0$ and $1$ (i.e. it is a Bernoulli variable), then $2I_1$ can take the values $0$ and $2.$ However $I_1+I_2$ can take the value $0$ (if both are zero), $1$ (if one is zero and the other is one), or $2$ (if both are one).

share|cite|improve this answer

answered Jul 17 at 1:19

spaceisdarkgreen

27.6k21547

$S$ does not have the same distribution as $nI_1.$ In fact, you have proven this by showing they have different variances. As an example, if $I_1$ can take the values $0$ and $1$ (i.e. it is a Bernoulli variable), then $2I_1$ can take the values $0$ and $2.$ However $I_1+I_2$ can take the value $0$ (if both are zero), $1$ (if one is zero and the other is one), or $2$ (if both are one).

share|cite|improve this answer

answered Jul 17 at 1:19

spaceisdarkgreen

27.6k21547

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 1:19

spaceisdarkgreen

27.6k21547

answered Jul 17 at 1:19

spaceisdarkgreen

27.6k21547

answered Jul 17 at 1:19

spaceisdarkgreen

27.6k21547

27.6k21547

add a comment | 

add a comment | 

up vote
6
down vote

You're missing the difference between these two things:
$$
beginalign
& I_1 + I_2+ cdots + I_n, \[8pt] & I_1 + I_1+ cdots + I_1.
endalign
$$
In one of those, the terms are independent; in the other, they're as far from independent as they can get.

NOTE: These do not both have the same distribution.

share|cite|improve this answer

answered Jul 17 at 3:51

Michael Hardy

204k23186462

add a comment | 

up vote
6
down vote

You're missing the difference between these two things:
$$
beginalign
& I_1 + I_2+ cdots + I_n, \[8pt] & I_1 + I_1+ cdots + I_1.
endalign
$$
In one of those, the terms are independent; in the other, they're as far from independent as they can get.

NOTE: These do not both have the same distribution.

share|cite|improve this answer

answered Jul 17 at 3:51

Michael Hardy

204k23186462

add a comment | 

up vote
6
down vote

up vote
6
down vote

You're missing the difference between these two things:
$$
beginalign
& I_1 + I_2+ cdots + I_n, \[8pt] & I_1 + I_1+ cdots + I_1.
endalign
$$
In one of those, the terms are independent; in the other, they're as far from independent as they can get.

NOTE: These do not both have the same distribution.

share|cite|improve this answer

answered Jul 17 at 3:51

Michael Hardy

204k23186462

You're missing the difference between these two things:
$$
beginalign
& I_1 + I_2+ cdots + I_n, \[8pt] & I_1 + I_1+ cdots + I_1.
endalign
$$
In one of those, the terms are independent; in the other, they're as far from independent as they can get.

NOTE: These do not both have the same distribution.

share|cite|improve this answer

answered Jul 17 at 3:51

Michael Hardy

204k23186462

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 3:51

Michael Hardy

204k23186462

answered Jul 17 at 3:51

Michael Hardy

204k23186462

answered Jul 17 at 3:51

Michael Hardy

204k23186462

204k23186462

add a comment | 

add a comment | 

up vote
4
down vote

We can not say $S$ has the same distribution as $nI_1$.

Example:

Toss two coins and count the heads as $I_1,I_2$ and their sum as $S$.   Now $I_1in0,1$, as is $I_2$, and they are iid as required.   So $Sin 0,1,2$ and yet $2I_1in 0,2$.   Thus we see that $S$ and $2I_1$ do not have the same support, therefore they cannot have the same distribution.

share|cite|improve this answer

answered Jul 17 at 1:28

Graham Kemp

80.1k43275

add a comment | 

up vote
4
down vote

We can not say $S$ has the same distribution as $nI_1$.

Example:

Toss two coins and count the heads as $I_1,I_2$ and their sum as $S$.   Now $I_1in0,1$, as is $I_2$, and they are iid as required.   So $Sin 0,1,2$ and yet $2I_1in 0,2$.   Thus we see that $S$ and $2I_1$ do not have the same support, therefore they cannot have the same distribution.

share|cite|improve this answer

answered Jul 17 at 1:28

Graham Kemp

80.1k43275

add a comment | 

up vote
4
down vote

up vote
4
down vote

We can not say $S$ has the same distribution as $nI_1$.

Example:

Toss two coins and count the heads as $I_1,I_2$ and their sum as $S$.   Now $I_1in0,1$, as is $I_2$, and they are iid as required.   So $Sin 0,1,2$ and yet $2I_1in 0,2$.   Thus we see that $S$ and $2I_1$ do not have the same support, therefore they cannot have the same distribution.

share|cite|improve this answer

answered Jul 17 at 1:28

Graham Kemp

80.1k43275

We can not say $S$ has the same distribution as $nI_1$.

Example:

Toss two coins and count the heads as $I_1,I_2$ and their sum as $S$.   Now $I_1in0,1$, as is $I_2$, and they are iid as required.   So $Sin 0,1,2$ and yet $2I_1in 0,2$.   Thus we see that $S$ and $2I_1$ do not have the same support, therefore they cannot have the same distribution.

share|cite|improve this answer

answered Jul 17 at 1:28

Graham Kemp

80.1k43275

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 1:28

Graham Kemp

80.1k43275

answered Jul 17 at 1:28

Graham Kemp

80.1k43275

answered Jul 17 at 1:28

Graham Kemp

80.1k43275

80.1k43275

add a comment | 

add a comment | 

 

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