Let $X$ and $Y$ be independent uniformly distributed random variables with $\mathbb{E}X=\mathbb{E}Y=0$ and $\mathbb{D}X=\mathbb{D}Y=4$. The task is to find following probability: $$P(4\leq (X^2+Y^2) \leq 9)$$ P.S: Here $\mathbb{D}$ stands for variance.
I've tried the following:
Let's find $\varphi_{X^2+Y^2}(t)=(\varphi_X^2(t))^2$ $\varphi_X^2(t)=...=\cfrac{1}{\sqrt{8-it}}$, so $\varphi_{X^2+Y^2}(t)=\cfrac{1}{1-8it}$, and that leads us to: $X^2+Y^2 \sim \text{Exp}(\frac{1}{8})$.
And so on. However I was told that this solution is not correct and there is more efficient and smart one. May you help me?
Probably $\mathbb{D}X$ is the variance.
The problem is very simple. If you do a drawing, the requested probability is the probability of an annulus resulting as the difference of the two disks, the greater with radius $3$ and the inner with radius $2$
Knowing mean and variance you can calculate $f_X$ and $f_Y$
Thus the solution is simply the purple area multiplied by the joint density or, what is the same, the purple area divided by the total area of the square.
If I assumed correctly $\mathbb{D}$ as Variance, easy find that
$$f_X(x)=\frac{1}{4\sqrt{3}}\cdot\mathbb{1}_{(-2\sqrt{3};2\sqrt{3})}(x)$$
thus the requested probability is
$$\mathbb{P}[4\leq X^2+Y^2\leq 9]=\frac{1}{16\cdot3}(9-4)\pi\approx 32.72\%$$