Var(x) = E[ x^2] - (E[X])^

[vptran84]

Hi,

Can someone please please show me why Var(x) = E[ x^2] - (E[X])^2.

I just dont get it. THanks in advance.

 

[kleinwolf]

Var(X)=E([X-E(X)]^2)

 

[vptran84]

yeah i got that part, similar to distributive property, but where does variance come from? Like how did they get E[x^2]-(E[X])^2 ??? How did they get E[x^2] and (E[X])^2 ??

 

[SpaceTiger]

yeah i got that part, similar to distributive property, but where does variance come from? Like how did they get E[x^2]-(E[X])^2 ??? How did they get E[x^2] and (E[X])^2 ??

$$(x-<x>)^2=x^2-2x<x>+<x>^2$$

$$\sigma^2=<(x-<x>)^2>=<x^2-2x<x>+<x>^2>=<x^2>-2<x><x>+<x>^2$$

$$\sigma^2=<x^2>-<x>^2$$

This just follows from the properties of expectation values, which follow from the properties of integrals:

$$<x>=\frac{\int xf(x)dx}{\int f(x)dx}=\int xP(x)dx$$

 

[BicycleTree]

$$<x>=\frac{\int xf(x)dx}{\int f(x)dx}=\int xP(x)dx$$

So <x> = E(x). The denominator there, $$\int f(x)dx$$, always equals 1 because f is a probability density function. The definition of an expected value is just the numerator of that fraction.

 

[SpaceTiger]

So <x> = E(x). The denominator there, $$\int f(x)dx$$, always equals 1 because f is a probability density function. The definition of an expected value is just the numerator of that fraction.

I'm not defining f(x) to be the probability density, just some distribution function. P(x) is the probability density. Sorry for not making that clear. But yes, <x>=E(x).

 

[BicycleTree]

So then what is <x>? Are you just defining it here or does it mean something else?

What's the difference between a distribution function and a density function? In my course they were used synonymously, except a distribution function could also be a probability mass function.

 

[SpaceTiger]

So then what is <x>?

You were right, it's the expectation value.

What's the difference between a distribution function and a density function?

I suppose I was using the wrong terminology. What my advisors sometimes call simply a "distribution function" is sometimes actually a "frequency distribution". This is what I meant by f(x). The idea is that the integral over its domain is not equal to one, but is instead equal to the number of objects in the sample (for example). The "distribution function" is actually something entirely different; that is, the cumulative probability of the value being less than x. Check Mathworld for the definitions of these things if you want more precision. If you don't want to bother (I wouldn't blame you), then disregard the middle part of my last equation (with f(x)) and just consider the part with P(x), the probability density.

 

[BicycleTree]

Yes, I know what a distribution function is.

It's just a different notation. In the course I took, P(...) means the probability of the stuff in the parentheses (which would be a logical formula). So you might say P(X=x). Also, distribution functions were denoted by capital letters and density/mass functions were denoted by the corresponding lowercase letters, so even if P didn't mean "the probability of," it would be a distribution function, not a density function.

 

[jetoso]

Answer

Hi,

Can someone please please show me why Var(x) = E[ x^2] - (E[X])^2.

I just dont get it. THanks in advance.

For a random variable X, the variance is defined as: Var(X) = E[(X-E[X])^2].
Thus, Var(X) = E[(X^2 - 2XE[X] - (E[X])^2]. Remember that the expected value of a constant, say a, is this constant: E(a) = a, where a is a constant. And also that: E[ E[X] ] = E[X].
Then, we have that:
Var(X) = E[X^2] - E[2XE[X]] - E[(E[X])^2] = E[X^2] - 2E[X]E[X] - (E[X])^2
Var(X) = E[X^2] - 2(E[X])^2 - (E[X])^2 = E[X^2] - (E[X])^2

Var(X) = E[X^2] - (E[X])^2, which is what you are trying to prove.