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

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Cov(Z, W) is calculated using the definitions of Z and W, leading to Cov(Z, W) = Var(X) + E(X^2)E(Y^2) - E(X)²E(Y^2). Given that X is a standard normal variable, Var(X) equals 1 and E(X²) also equals 1, resulting in Cov(Z, W) = 2. The discussion also clarifies that the variance of X is derived from the expected value and its square, confirming E(X²) = 1. Additionally, it is noted that if Y = X², then Y follows a chi-squared distribution, not a normal distribution.
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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)?
 
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Re: E(x^2)and VAR(x^2)if X is N(0,1).

Dhamnekar Winod said:
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:
$$\operatorname{Var}(X) = E(X^2) - (EX)^2$$

Dhamnekar Winod said:
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:
$$\operatorname{Var}(X) = E(X^2) - (EX)^2 \implies 1=E(X^2)-0 \implies E(X^2)=1$$
 
Re: E(x^2)and VAR(x^2)if X is N(0,1).

Klaas van Aarsen said:
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?
 
Re: E(x^2)and VAR(x^2)if X is N(0,1).

Dhamnekar Winod said:
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.
 
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