X and Y are independent standard normal random variables, which is of the following isn't true?

Question

Which of the following is not true?

$X-Y$ and $X + Y$ are normally distributed.
$X-Y$ and $X + Y$ are independent.
$E[X^2Y^2] = 1$
$E[X^2 / Y^2] = 1$

I'm confused. All of these look true to me. 1 is true because any linear combination of independent random normals is normally distributed. 2 is true from Are the random variables $X + Y$ and $X - Y$ independent if $X, Y$ are distributed normal?. 3 is true because $X,Y$ are independent, so we have $E[X^2Y^2] = E[X^2]E[Y^2]$. We know $E[X^2] = E[Y^2] = 1$ so $E[X^2]E[Y^2] = 1$. 4 is true because $E[X^2 / Y^2] = E[X^2] / E[Y^2] = 1/1=1$.

Did I do something wrong here or are all 4 options true?

score 5 · Accepted Answer · answered Aug 09 '21 at 19:42

5

$\newcommand{\e}{\operatorname E}$We certainly do not know $\e\left( \dfrac{X^2}{Y^2} \right) = \dfrac{\e(X^2)}{\e(Y^2)}.$

One can say that $\e\left( \dfrac{X^2}{Y^2} \right) = \e(X^2)\e\left( \dfrac 1 {Y^2} \right)$ (by independence), and then one must think about $\e\left( \dfrac 1 {Y^2} \right).$

answered Aug 09 '21 at 19:42

Michael Hardy

1

Ahhh. I see. Is there some property for $E[1 / Y^2]$ in this particular case? – student010101 Aug 09 '21 at 19:45
$$ \begin{align} & \int_{-\infty}^{+\infty} \frac 1 {y^2} \varphi(y),dy \ {} \ \ge {} & \int_{-1}^{+1} \frac 1 {y^2}, dy/10 \end{align} $$ since $\varphi(y)>1/10$ when $-1<y<1.$ Here $\varphi$ is the standard normal density. $\qquad$ – Michael Hardy Aug 09 '21 at 19:48
@student010101 : $\qquad\uparrow\qquad$ – Michael Hardy Aug 16 '21 at 05:09

X and Y are independent standard normal random variables, which is of the following isn't true?

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