Csdrhrt

Suppose that X and Y are independent n(0,1) random variables.

a) Find $P(X^2+Y^2<1)$

Attempt:

a) Let $U = X^2 + Y^2$, $V = Y$.

Then $X = sqrt{V^2 -U}$, $Y = V$.

$J = left| begin{array}{ccc}
frac{-1}{sqrt{V^2-U}} & frac{V}{V^2-U} \
0 & 1\
end{array} right| $

Then the joint distribution of $f_{u,v}(u,v)$ is:

$$f_{u,v}(u,v)= frac{1}{2pi}e^{frac{-sqrt{v^2-u}}{2}}e^{frac{-u^2}{2}}frac{1}{sqrt{v^2-u}}$$

Then $P(X^2 +Y^2 <1)$ is:

$$int_0^infty int_0^{v^2-u} frac{1}{2pi}e^{frac{-sqrt{v^2-u}}{2}}e^{frac{-u^2}{2}}frac{1}{sqrt{v^2-u}}dudv$$

However, at this point I simply do not know how any tricks to complete this integration.

Since $X$ and $Y$ are independent, the joint density of $X$ and $Y$ is the product of the individual densities, so it is
$$frac{1}{2pi}e^{-(x^2+y^2)/2}$$
on the entire $x$-$y$ plane. We want to find
$$int_D frac{x^2+y^2}{2pi}e^{-(x^2+y^2)/2},dy,dx,$$
where $D$ is the unit disk. Switch to polar coordinates. We want to find
$$int_{r=0}^{1}int_{theta=0}^{2pi} frac{r^3}{2pi}e^{-r^2/2},dtheta,dr.$$
The integration with respect to $theta$ gets rid of the $2pi$ in the denominator.
So we want
$$int_0^1 r^3 e^{-r^2/2},dr.$$
This can be done by integration by parts. To make things more familiar, you may wish to first make the substitution $t=r^2/2$. Our integral becomes
$$int_0^{1/2} 2te^{-t},dt.$$

Although the following method is more convoluted than necessary, I think it provides a fun way to see the convolution method for computing the density of a sum of independent random variables in action. It also requires a few fun integration tricks.

First we will find the distribution for $X^2$ (and, by symmetry, the distribution of $Y^2$). We consider the cdf $F_{X^2}(t)$, and we see that $$F_{X^2}(t) = P(X^2 leq t) = P(|X| leq sqrt{t}) = P(X in [-sqrt{t},sqrt{t}] = P(X leq sqrt{t}) - P(X < sqrt{t}) $$ Continuing this logic, and noting that $X$ has density $f_X = frac{1}{sqrt{2 pi}} e^{-x^2/2}$, we see that $$ F_{X^2}(t) = F_X(sqrt{t}) - F_X(-sqrt{t}) = displaystyleint_{-infty}^{sqrt{t}} frac{1}{sqrt{2 pi}} e^{-x^2/2} dx - displaystyleint_{-infty} ^{-sqrt{t}} frac{1}{sqrt{2 pi}} e^{-x^2/2} dx$$ We can then find the density $f_{X^2}$ by differentiating our equation for $F_{X^2}(t)$ with respect to $t$, which gives us $$f_{X^2}(t) = frac{d}{dt}left(F_{X^2}(t) right) = frac{1}{sqrt{2 pi}} e^{- (sqrt{t})^2 / 2} left( frac{1}{2 sqrt{t}} right) - frac{1}{sqrt{2 pi}} e^{- (-sqrt{t})^2 / 2} left(- frac{1}{2 sqrt{t}} right) $$ and this simplifies to tell us that $$f_{X^2}(t) = f_{Y^2}(t) = frac{1}{sqrt{t} sqrt{2 pi}} e^{- t/2} $$

Because we know that $X$ and $Y$ are independent, it follows that $X^2$ and $Y^2$ are independent as well. Therefore we can apply the convolution method to find the probability density for $Z = X^2 + Y^2$. We can write the probability distribution as the convolution $$f_Z(z) = displaystyleint_{0}^z f_{X^2}(z-y) f_{Y^2}(y) dy $$ where the lower-limit of the integral is 0 because $Y^2$ is a non-negative random variable, and the upper limit is $z$ because it cannot be the case that both $Y^2 = y > z$ and $X^2 < z - y$, as $X^2 $ is a non-negative random variable. Plugging our value for the densities $f_{X^2}(z-y) = frac{1}{sqrt{z-y} sqrt{2 pi}} e^{-(z-y)/2} $ and $f_{Y^2} = frac{1}{sqrt{y} sqrt{2 pi}} e^{-y/2}$, we have that $$f_Z(z) = displaystyleint_0^z left( frac{1}{sqrt{z-y} sqrt{2 pi}} e^{-(z-y)/2} right) left( frac{1}{sqrt{y} sqrt{2 pi}} e^{-y/2} right) dy = frac{1}{2 pi} e^{-z/2} displaystyleint_0^z displaystylefrac{dy}{sqrt{zy - y^2}}$$

Now, what we have left to do is to compute the integral on the right. Completing the square under the square root sign we have $$displaystyleint_0^z displaystylefrac{dy}{sqrt{zy - y^2}} = displaystyleint_0^z displaystylefrac{dy}{sqrt{frac{z^2}{4} - (frac{z^2}{4} - zy + y^2) }} = displaystyleint_0^z displaystylefrac{dy}{sqrt{frac{z^2}{4} - (y - frac{z}{2})^2 }} $$ Making the u-substitution $u = y - frac{z}{2}$ (with $du = dy$ and with transformed limits of integration $y = 0 mapsto u = -z/2$ and $y = z mapsto z/2$), we obtain the integral $$ displaystyleint_0^z displaystylefrac{dy}{sqrt{zy - y^2}} = displaystyleint_{-z/2}^{z/2} displaystylefrac{du}{sqrt{frac{z^2}{4} - u^2}} = frac{2}{z} displaystyleint_{-z/2}^{z/2} displaystylefrac{du}{sqrt{1 - (frac{2u}{z})^2}} = frac{2}{z} left( left(frac{1}{frac{2}{z}} right) sin^{-1}(frac{2u}{z}) bigg|_{-z/2}^{z/2} right) $$ This simplifies to $$ displaystyleint_0^z displaystylefrac{dy}{sqrt{zy - y^2}} = sin^{-1}(1) - sin^{-1}(-1) = frac{pi}{2} - (- frac{pi}{2}) = pi $$ and therefore we have that $$f_{Z}(z) = frac{1}{2 pi} e^{-z/2} left( pi right) Longrightarrow boxed{f_{Z}(z) = frac{1}{2} e^{-z/2}} $$

And we can compute that $$P(X^2 + Y^2 <1 ) = P(Z < 1) = displaystyleint_0^1 f_{Z}(z)dz = displaystyleint_0^1 frac{1}{2} e^{-z/2} dz = - e^{-z/2} bigg|_0^1 $$ and we obtain our final answer $$ boxed{P(X^2 + Y^2 < 1) = 1 - e^{-1/2}} $$

Since $X$ and $Y$ are independent, and because $mu=0$ and $sigma=1$, then we know that $$f_{X,Y}(x,y)=f_X(x)f_Y(y)=frac{1}{2pi}exp(frac{-(x^2+y^2}{2}).$$

Of course, since our domain over which we are integrating is a circle with a radius of 1, we convert $x^2+y^2=r^2$, and $text{d}x text{d}y=rtext{d}rtext{d}theta$, using
$x=rcostheta$ and $y=rsin theta$ using the standard polar coordinate transformation. Notice, we are converting the joint pdf $f_{X,Y}(x,y)$ to make our lives easier, and the circle defined by the question, ie $X^2+Y^2<1$, merely comes into play when looking at what we integrate over.

begin{align*}
P(X^2+Y^2<1)&=frac{1}{2pi}int_{r=0}^{1}int_{theta=0}^{2pi}e^{(frac{-r^2}{2})}r text{d}r text{d}theta\
&=2pi frac{1}{2pi}int_{r=0}^{1}e^{(frac{-r^2}{2})}r text{d}r\
&=int_{u=1}^{e^{-1/2}}-text{d}underbrace{(e^{-r^2/2})}_u=-ubigg|_1^{e^{-1/2}}\
&=-e^{1/2}--1=1-e^{-1/2}
end{align*}

As per my knowledge you don't need to integrate , this type of questions are normally asked in competetive exam if you started integrating there then it will take much time. I have a shortcut.

$$Xsim N(0,1)$$

$$Y sim N(0,1)$$

then

$$X^2 simchi_1^2$$

$$Y^2 simchi_1^2$$

$$chi_1^2+chi_1^2 simchi_2^2$$

Chi-square distribution with $2$ degree of freedom will follow Gamma distribution with parameters $(1,frac12)$ which is an exponential distribution with parameter $lambda=frac12$ so now we have an exponetial distribution lets call it $T$ so

$$ T sim expo(1/2) $$

we have to find

$$P(T<1)$$

as we know $P(T<1)$ is cumulative ditribution function of exponetial ditribution which is generally given by

$$F(x) = 1- e^{-lambda x}$$

as we know $lambda$=1/2
$$F(1)= 1-e^{{-{1over 2}}times 1}$$

$$Rightarrow P(X^2 + Y^2 < 1) = 1-e^{-1/2}$$

Box-Muller transform suggests that if $X, Y$ are two independent standard normal random variables, then if we let $X = R cosTheta, Y = R sinTheta, R in (0, infty), Theta in [0, 2pi)$ (polar coordinate), then this bivariate transform is 1-to-1 and onto. Further, $R, Theta$ are also independent. And $frac{1}{2}R^2 sim Exponential(1), Theta sim Uniform(0, 2pi)$, i.e., $frac{1}{2}R^2$ follows an Exponential distribution with $lambda = 1$ and $Theta$ follows a Uniform distribution on $(0, 2pi)$.

Hence, $P(X^2 + Y^2 < 1) = P(R^2cos^2Theta + R^2sin^2Theta < 1) = P(R^2 < 1) = P(frac{1}{2}R^2 < frac{1}{2})$. And since $frac{1}{2}R^2 sim Exponential(1)$, we can denote $Z = frac{1}{2}R^2$. Using the pdf of an $Exponential(1)$ distribution, $P(frac{1}{2}R^2 < frac{1}{2}) = P(Z < frac{1}{2}) = int_{-infty}^{frac{1}{2}} e^{-z} , dz = 1 - e^{-frac{1}{2}}$.