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Q. Bullets are fired at the origin of an $(x,y)$ coordinate system, and point hit, say $(X,Y)$ is a random variable. The Random variables $X,Y$ follow standard Normal. If two bullets are fired independently, what is the distribution of the distance between the random variables.

Attempt: I tried find the joint pdf of $X,Y$ then I noticed that if the points are $(X_1,Y_1)$ and $(X_2,Y_2)$ then we need to find the distribution of ${((X_1-X_2)^2+(Y_1-Y_2)^2)}^{1/2}$

Notice that the following ${(X_1-X_2)^2+(Y_1-Y_2)^2)}$ follows Chi Squared distribution with degrees of freedom$=2$

Now trying to find the pdf of the square root independently and thereby finding the distribution of the whole is becoming very tedious, does there exist any simpler way to solve this.

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If $$X, Y \sim N(0,1),$$

you have

$$X_1-X_2\sim N(0,2), \qquad Y_1-Y_2\sim N(0,2)$$

and hence

$$((X_1-X_2)^2/2 + (Y_1-Y_2)^2/2) \sim \chi^2(2).$$

Therefore, you get

$$ \sqrt{(X_1-X_2)^2+ (Y_1-Y_2)^2} \sim \sqrt{2}\,\chi(2).$$

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We need the distribution of $R=\sqrt{(X_1-X_2)^2+(Y_1-Y_2)^2}.$ Since $X_1,X_2,Y_1,Y_2$ are all independent standard normal random variables, we have $Z_1=X_1-X_2\sim \mathrm{N}(0,2)$ and similarly, $Z_2=Y_1-Y_2\sim \mathrm{N}(0,2).$ Note that $Z_1,Z_2$ are independent. As a result, $$U=\dfrac{Z_1^2+Z_2^2}{2}=\dfrac{R^2}{2}\sim \chi^2_2.$$

We can now find the distribution of $R$ using transformation of random variables. The transformation is $R=\sqrt{2U}.$ The pdf of $U$ is $$f_U(u)=\dfrac{e^{-u/2}}{2}\hspace{1cm}\text{ for }u>0. $$

That means the pdf of $R$ should be $f_R(r)=\dfrac{re^{-r^2/4}}{2}$ for $r>0$.

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