Find Cov(Z,W) with E(X^2)if X is N(0,1)

[WMDhamnekar]

Let X and Y be two independent $$\mathcal{N}(0,1)$$ random variables and

$$Z=1+X+XY^2$$

$$W=1+X$$
I want to find Cov(Z,W).

Solution:-

$$Cov(Z,W)=Cov(1+X+XY^2,1+X)$$

$$Cov(Z,W)=Cov(X+XY^2,X)$$

$$Cov(Z,W)=Cov(X,X)+Cov(XY^2,X)$$

$$Cov(Z,W)=Var(X)+E(X^2Y^2)-E(XY^2)E(X)$$

$$Cov(Z,W)=1+E(X^2)E(Y^2)-E(X)^2E(Y^2)$$

$$Cov(Z,W)=1+1-0=2$$

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

 

[I like Serena]

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

My another question is what is $$Var(X^2)?$$

Hi Dhamnekar,

Let's start with this one.
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 $$\mathcal{N}(0,1)$$ random variables and

(snip)

Now E(X)=0, So $$E(X)^2E(Y^2)=0$$, But i don't follow how $$E(X^2)E(Y^2)=1?$$ 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:
$$\operatorname{Var}(X) = E(X^2) - (EX)^2 \implies 1=E(X^2)-0 \implies E(X^2)=1$$

 

[WMDhamnekar]

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

Hi Dhamnekar,

Let's start with this one.
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?

 

[I like Serena]

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.​