# E(x^2)if X is N(0,1).

#### Dhamnekar Winod

##### Active member
Let X and Y be two independent [Tex]\mathcal{N}(0,1)[/Tex] random variables and

[Tex] Z=1+X+XY^2[/Tex]

[Tex]W=1+X[/Tex]
I want to find Cov(Z,W).

Solution:-

[Tex]Cov(Z,W)=Cov(1+X+XY^2,1+X)[/Tex]

[Tex]Cov(Z,W)=Cov(X+XY^2,X)[/Tex]

[Tex]Cov(Z,W)=Cov(X,X)+Cov(XY^2,X)[/Tex]

[Tex]Cov(Z,W)=Var(X)+E(X^2Y^2)-E(XY^2)E(X)[/Tex]

[Tex]Cov(Z,W)=1+E(X^2)E(Y^2)-E(X)^2E(Y^2)[/Tex]

[Tex]Cov(Z,W)=1+1-0=2[/Tex]

Now E(X)=0, So [Tex]E(X)^2E(Y^2)=0[/Tex], But i don't follow how [Tex]E(X^2)E(Y^2)=1?[/Tex] Would any member explain that? My another question is what is [Tex]Var(X^2)?[/Tex]

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: E(x^2)and VAR(x^2)if X is N(0,1).

My another question is what is [Tex]Var(X^2)?[/Tex]
Hi Dhamnekar,

It's the variance. It is the mean of the squared deviations from the average.
And the average of $X$ is the same thing as the expected value $E(X)$ or just $EX$.
In formula form:
$$\operatorname{Var}(X) = E\left((X - EX)^2\right)$$
If we write it out, we can find that it can be rewritten as:
$$\operatorname{Var}(X) = E(X^2) - (EX)^2$$

Let X and Y be two independent [Tex]\mathcal{N}(0,1)[/Tex] random variables and

(snip)

Now E(X)=0, So [Tex]E(X)^2E(Y^2)=0[/Tex], But i don't follow how [Tex]E(X^2)E(Y^2)=1?[/Tex] Would any member explain that?
Now let's get back to your first question.

The fact that $X \sim \mathcal{N}(0,1)$ means that $\operatorname{Var}(X)=1$.
Combine it with $EX=0$ and fill it in:
$$\operatorname{Var}(X) = E(X^2) - (EX)^2 \implies 1=E(X^2)-0 \implies E(X^2)=1$$

#### Dhamnekar Winod

##### Active member
Re: E(x^2)and VAR(x^2)if X is N(0,1).

Hi Dhamnekar,

It's the variance. It is the mean of the squared deviations from the average.
And the average of $X$ is the same thing as the expected value $E(X)$ or just $EX$.
In formula form:
$$\operatorname{Var}(X) = E\left((X - EX)^2\right)$$
If we write it out, we can find that it can be rewritten as:
$$\operatorname{Var}(X) = E(X^2) - (EX)^2$$

Now let's get back to your first question.

The fact that $X \sim \mathcal{N}(0,1)$ means that $\operatorname{Var}(X)=1$.
Combine it with $EX=0$ and fill it in:
$$\operatorname{Var}(X) = E(X^2) - (EX)^2 \implies 1=E(X^2)-0 \implies E(X^2)=1$$
Hello,
If $X$ be $\mathcal{N}(0,1)$ random variable, and $Y=X^2$ is the function of $X$, what is the distribution of $Y$?

Is its distribution Normal?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: E(x^2)and VAR(x^2)if X is N(0,1).

Hello,
If $X$ be $\mathcal{N}(0,1)$ random variable, and $Y=X^2$ is the function of $X$, what is the distribution of $Y$?

Is its distribution Normal?
No...

In probability theory and statistics, the chi-squared distribution (also chi-square or $χ^2$-distribution) with $k$ degrees of freedom is the distribution of a sum of the squares of $k$ independent standard normal random variables.