Distribution of $\Big(Y_1+Y_2\Big)^2$ and $\Big(Y_1-Y_2\Big)^2$ where $Y_i \sim N(0,1)$

Question

Does anyone know what is the distribution of $(Y_1+Y_2)^2$ and $(Y_1-Y_2)^2$ where $Y_i \sim N(0,1)$ are independent variables? I have tried to go through the joint pdf, but when trying to change variables I have multiples cases because it is squared.

Answer 1

$(Y_1+Y_2)\sim N(0;2)$

$\frac{Y_1+Y_2}{\sqrt{2}}\sim N(0;1)$

$(\frac{Y_1+Y_2}{\sqrt{2}})^2\sim \chi_{(1)}^2=Gamma(\frac{1}{2};\frac{1}{2})$

$(Y_1+Y_2)^2 \sim 2Gamma(\frac{1}{2};\frac{1}{2})=Gamma(\frac{1}{2};\frac{1}{4})$

Show you efforts for the second case

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Here are the detailed calculations

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Let's $Z=\frac{Y_1+Y_2}{\sqrt{2}}$

applying properties of Gaussian distribution $Z \sim N(0;1)$

Deriving the Law of $W=Z^2$ via Fundamental Transformation Theorem you get

$f_W(w)=\frac{1}{\sqrt{2\pi}}e^{-\frac{y}{2}}\frac{1}{2\sqrt{y}}+\frac{1}{\sqrt{2\pi}}e^{-\frac{y}{2}}\frac{1}{2\sqrt{y}}$

Now re-organizing $f(w)$ we have

$f_W(w)=\frac{(\frac{1}{2})^{\frac{1}{2}}}{\sqrt{\pi}}y^{-\frac{1}{2}}e^{-\frac{y}{2}}=\frac{(\frac{1}{2})^{\frac{1}{2}}}{\Gamma(\frac{1}{2})}y^{\frac{1}{2}-1}e^{-\frac{y}{2}}\sim Gamma(\frac{1}{2};\frac{1}{2})$

(also known as $\chi_{(1)}^2$)

Multiplying this density by 2 you get again a Gamma, a $Gamma(\frac{1}{2};\frac{1}{4})$

This result can be easily derived again via Fundamental Transformation Theorem

Answer 2

Hint: $Z=\frac {Y_1+Y_2} {\sqrt 2} \sim N(0,1)$ assuming independence of $Y_1$ and $Y_2$. Hence $(Y_1+Y_2)^{2} \sim 2 Z^{2}$. Same answer for $(Y_1-Y_2)^{2}$.

$Z^{2}$ has chi-square distribution with one degree of freedom.

Answer 3

What are you allowed/expected to do? Some of these derivations are not that easy. For example, using the CDF to derive PDF of $Z=Y_1 + Y_2$, where $Y_1, Y_2 \sim N(0,1)$ goes like this ($\Phi$ and $\varphi$ are CDF and PDF of standard normal rv): $$ F_Z(z) = P(Z< z) = P(X_1 + X_2 < z) = \int_{-\infty}^{\infty} P(X_1 < z-x_2|X_2=x_2)\varphi(x_2)dx_2\\ = \int_{-\infty}^{\infty}\Phi(z-x_2) \varphi(x_2)d x_2 $$ The third step is due to independence of rvs. Now, differentiating $F(z)$ to get $f(z)$: $$ f(z) = \int_{-\infty}^{\infty} \varphi(z-x_2)\varphi(x_2) dx_2 = \int_{-\infty}^{\infty}\frac{1}{\sqrt{2 \pi}} e^{-\frac{(z-x_2)^2}{2}}\frac{1}{\sqrt{2 \pi}} e^{-\frac{x_2^2}{2}} dx_2 \\ = \frac{1}{\sqrt{2 \pi}}\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-\frac{\frac{z^2}{2} + \frac{z^2}{2} - 2 \times\sqrt{2}x_2 \times \frac{z}{\sqrt{2}} + (\sqrt{2}x_2)^2}{2}} dx_2 \\ = \frac{1}{\sqrt{2 \pi}} e^{-\frac{z^2}{4}} \times \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{- \frac{\bigg(\frac{z}{\sqrt{2}} - \sqrt{2} x_2 \bigg)^2}{2}} dx_2 = \frac{1}{\sqrt{2 \pi}} \frac{1}{\sqrt{(1+1)}} e^{-\frac{z^2}{2 \times(1+1)}} $$ Since $\sigma_1^2 = \sigma_2^2 = 1, Z \sim N (0, 2)$.

You other problems should be also solved in a rigorous way.