Random walk - why is the STD equal sqrt(n)

[gony rosenman]

typical random walk :
one step forward or backward with equal probability and independence of each step , what is the expectation and Variance .

so i define indicator variable x_i ={1 or -1 with equal probability .
E(x_i) = 0
Var(x_i) = 1

now define S_n as the sum of i=1,...,n
each step is independent and identical to the other so
i can say that S_n = nx_i
now from definition of Expectation and Variance :
E(S_n) = E(nx_i) = nE(x_i) = 0
Var(S_n) = Var(nx_i) = n²Var(x_i) = n² ⇒σ_sn=n
but i know for certain that actually the answer is σ_sn=sqrt(n)
how is that ?
please
thank you!

 

[.Scott]

$S_n$ is the sum of $x_i: i=1..n$
That is not $nx_i$.

$S_n$ would be {$-n, ... , +n$} with a binomial distribution.

 

[gony rosenman]

$S_n$ is the sum of $x_i: i=1..n$
That is not $nx_i$.

$S_n$ would be {$-n, ... , 0, ..., +n$} with a binomial distribution.

could you elaborate further?

 

[.Scott]

could you elaborate further?

For n=4:
$Sn$ would be {-4, p=1/16; -2, p=4/16; 0, p=6/16; +2, p=4/16; +4, p=1/16}
or {-4, -2, -2, -2, -2, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 4}

Actually, in post #2, that zero I put in the middle is wrong (I'll edit it):
For n=5:
$Sn$ would be {-5, p=1/32; -3, p=5/32; -1, p=10/16; +1, p=10/32; +3, p=5/32; +5, p=1/32}
or {-5, -3, -3, -3, -3, -3, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 5}

 

[gony rosenman]

For n=4:
$Sn$ would be {-4, p=1/16; -2, p=4/16; 0, p=6/16; +2, p=4/16; +4, p=1/16}
or {-4, -2, -2, -2, -2, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 4}

Actually, in post #2, that zero I put in the middle is wrong (I'll edit it):
For n=5:
$Sn$ would be {-5, p=1/32; -3, p=5/32; -1, p=10/16; +1, p=10/32; +3, p=5/32; +5, p=1/32}
or {-5, -3, -3, -3, -3, -3, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 5}

yes but how do you show that σ=√n

 

[.Scott]

I was just showing that your logic was wrong. I wasn't computing what it would actually be.

 

[gony rosenman]

I was just showing that your logic was wrong. I wasn't computing what it would actually be.

are you sure you are capable to assist me ?
i think you might not know what an indicator random variable is and thus you confuse me and might cause actual damage , even though I'm sure your intentions are good

 

[gony rosenman]

o.k so i figured it out by myself , sor anyone who might stumble upon this confusion i will explain :
so s_n ≠ nx_i but actually s_n = ∑ⁿ_i=1 x_i
so E(s_n) = E(x₁) +...+ E(x_n) = 0
and Var(s_n) = Var(x₁) +...Var(x_n) = n ⇒σ_sn= √n
my mistake was for doing s_n = nx_i when actually it is s_n = ∑ⁿ_i=1 which is tremendously different .
good luck!

 

[.Scott]

I am using the same nomenclature that you introduced.
if $x_i$ = {$-1, 1$} with equal distribution
then the distribution you will get with $S_n$ will be what I showed.
When I take the population standard deviation of some example $S_n$'s, I get $\sqrt{n}$.

What is very likely is that your $nx_n$ is not normal multiplication. But if it isn't, you would be running into trouble when you use it as normal multiplication later in your work.

 

[Ray Vickson]

I am using the same nomenclature that you introduced.
if $x_i$ = {$-1, 1$} with equal distribution
then the distribution you will get with $S_n$ will be what I showed.
When I take the population standard deviation of some example $S_n$'s, I get $\sqrt{n}$.

What is very likely is that your $nx_n$ is not normal multiplication. But if it isn't, you would be running into trouble when you use it as normal multiplication later in your work.

What is more likely is that the OP originally made an error, and wrote down something without giving it a lot of thought.

In fact, the original argument of the OP is almost right: each $X_i$ has mean zero, so each has variance $\text{Var}(X_i) = E(X_i^2)$. However, since it really is true that all the $X_i^2 = 1$ are identical, it follows that the variance is $\sum 1 = n.$