I have been working on some exercises for probability. There is a problem that I cannot even figure out where to start. So, here is the question.

  • Let $T$ be drawn from a uniform distribution on the interval $\left[0, \,\sqrt{\,{2}\,}\, - 1\right]$.
  • Accept $T$ with probability $1/\left(1 + T^{2}\right)$, otherwise start over.
  • Let $S = 2T/\left(1 + T^{2}\right)$ and $C = 1 - ST$.
  • Now with probability $1/2$ switch $S$ and $C$.
  • Then with probability $1/2$ for each, independently, change the signs of $S$ and $C$.
  • What is the joint distribution of $S$ and $C$ ?.

Any comments would be appreciated. Thanks in advance !.


2 Answers 2


Okay, Let $ T \sim \mathcal U([0,\sqrt2-1]) $. Now, given that $T=t$ let $X$ be equal to $t$ with probability $\frac{1}{1+t^2}$. Otherwise, we select $T$ again. Firstly we'd like to find the distribution of $X$. Let $F_X(s) = \mathbb P(X \leq s)$. There are two trivial cases. If $s < 0 $ then $ F_X(s) = 0$ and if $s > \sqrt2 -1 $ then $F_X(s) = 1$ no matter what.

So now the hard one, let $s \in [0,\sqrt2-1] $. For $ \{X \leq s\} $ to happen, we must meet two conditions:

Obviously $T \leq s$ and while it happened, there also must be that $T$ was accepted.

$F_X(s) = \mathbb P(\{T\leq s\}|\{T "accepted"\}) = \mathbb P(\{T "accepted" \} \cap \{T\leq s \})\cdot \frac{1}{\mathbb P(T "accepted")} $.

So we have to compute probabilities of two events.

$\mathbb P(T "accepted") = \mathbb E[\chi_{_{T "accepted"}} ] = \mathbb E [ \mathbb E[\chi_{_{T "accepted"}} | T]] = \mathbb E[\frac{1}{1+T^2}] = \frac{1}{\sqrt2-1}\int_0^{\sqrt2-1}\frac{1}{1+t^2}dt = $

$ = \frac{\pi}{8(\sqrt2-1)}$, because $\arctan( \sqrt2-1) = \frac{\pi}{8} $

$\mathbb P(T \leq s \cap T "accepted" ) = \mathbb P(T"accepted" | T \leq s)\cdot \mathbb P(T\leq s) = \mathbb P(T"accepted" | T \leq s)\frac{s}{\sqrt2 -1 }$

That is $F_X(s) = \frac{8s}{\pi}\cdot \mathbb P(T"accepted" | T \leq s) $, while the latter is equal to $\frac{1}{s} \int_0^s \frac{1}{1+t^2}dt = \frac{\arctan(s)}{s}$ (because we have to normalize it over $[0,s]$ (so that we get a factor $\frac{1}{s}$) and then integrate that "conditional" probability over whole interval of possible values of $T$( note, it isn't the whole $[0,\sqrt2-1]$ now, but only $[0,s]$))

Which gives us $F_X(s) = \frac{8\arctan(s)}{\pi}$, for $s\in[0,\sqrt2-1]$, so that $g_X(s) = \frac{8}{\pi(1+s^2)}\chi_{[0,\sqrt2-1]}(s)$ is the density.

Now we have to deal with $ S = \frac{2X}{1+X^2} = f(X)$ and $ C = \frac{1-X^2}{1+X^2} = h(X)$.

Since $f,h$ are smooth functions we can (and will) use the rule $g_S(s) = g_X(f^{-1}(s))|\det(f^{-1}(s)|$ (similarly for C)

Let's attack $S$: $f([0,\sqrt2-1]) = [0,\frac{2\sqrt2-2}{2+2-2\sqrt2}] = [0,\frac{1}{\sqrt2}] $

$ Y = \frac{2X}{1+X^2} \iff Y + YX^2 = 2X \iff YX^2 - 2X + Y = 0 $

$\Delta = 4 - 4Y^2$ so $\sqrt\Delta = 2\sqrt{1-Y^2}$, and we see that $f^{-1}(y) = \frac{1\pm \sqrt{1-y^2}}{y}$ (but the one with plus sign has to be rejected due to or domain of $f^{-1}$ (if we allow plus sign, then $f^{-1}(\frac{1}{\sqrt{2}}) = \sqrt2 + 1$, which isn't good)).

So $f^{-1}(y) = \frac{1 - \sqrt{1-y^2}}{y}$, and it's derivative $\frac{\frac{2y^2}{2\sqrt{1-y^2}}-(1-\sqrt{1-y^2})}{y^2} = \frac{y^2 - \sqrt{1-y^2} + 1 - y^2}{y^2\sqrt{1-y^2}} = \frac{1-\sqrt{1-y^2}}{y^2\sqrt{1-y^2}} $

So let's plug it:

$g_S(s) = \frac{8}{\pi} \cdot \frac{1-\sqrt{1-y^2}}{y^2\sqrt{1-y^2}} \cdot \frac{y^2}{y^2 +2 - y^2 - 2\sqrt{1-y^2}} = \frac{4}{\pi\sqrt{1-y^2}}$, and thankfully it integrates to $1$ over $[0,\frac{1}{\sqrt2}] $

Doing similarly with $C$ now: $h([0,\sqrt2-1])=[\frac{1-(2+1-2\sqrt{2})}{1+(2+1-2\sqrt{2})},1] = [\frac{2\sqrt2-2}{4-2\sqrt2},1] = [\frac{1}{\sqrt2},1] $

Hmm, that can help in the future that $S$ and $C$ have different values.

Okay, but let's go, finding $h^{-1}$:

$ Z = \frac{1-X^2}{1+X^2} \iff Z + ZX^2 = 1 - X^2 \iff X^2(1+Z) = 1 - Z \iff X^2 = \frac{1-Z}{1+Z} $ And because we need positive value $ (X\ge 0 )$ so $X = \sqrt{\frac{1-Z}{1+Z}} $ and $h^{-1}(s) = \frac{\sqrt{1-s}}{\sqrt{1+s}}$ and we can calculate it's derivative : $\frac{-\sqrt{1+s}}{\sqrt{1-s}(1+s)^2}$, now plugging everything into $g_C$ we have:

$g_C(s) = \frac{8}{\pi}\cdot \frac{\sqrt{1+s}}{\sqrt{1-s}(1+s)^2} \cdot \frac{1+s}{1+s + 1 - s} = \frac{4}{\pi\sqrt{1-s^2}}$ and again integrates to $1$ over$[\frac{1}{\sqrt2},1]$

NOTE: Those densities equal to that value for $s \in [0,\frac{1}{\sqrt2}]$ and $s \in [\frac{1}{\sqrt2},1] $ respectivelly, otherwise they are zero.

So we should take them into one spot, because they can be helpful:

$g_S(x) = \frac{4}{\pi\sqrt{1-x^2}}\chi_{[0,\frac{1}{\sqrt2}]}(x)$, and $g_C(x) = \frac{4}{\pi\sqrt{1-x^2}}\chi_{[\frac{1}{\sqrt2},1]}(x)$

So now we arrived at the moment when with probability $\frac{1}{2}$ we have to swap those variables. Instead we will create two new variables, call them $W,U$, and we describe them as follow: when $C=c, S=s$ then $\mathbb P(W=s)=\mathbb P(W=c) = \frac{1}{2} $ and $U$ is always that remaining value (that is: when $W=s$ then $U=c$ and when $W=c$ then $U=s$). Okay, so now we have to independently choose signs for $W,U$. Anyway, we need joint distribution of $W$ and $U$ that is the distribution of random vector $V = (W,U) $ .We are still conditioning on $S=s, C=c$ (let $L=(S,C)$):

$\mathbb P(V=(\pm s,\pm c) | L=(s,c)) = \mathbb P(V=(\pm c,\pm s) | L=(s,c)) = \frac{1}{8} $ (Hope you understand my shortcuts)

Now, when $X=x$ that mean : $s=\frac{2x}{1+x^2}, c=\frac{1-x^2}{1+x^2} $ while $x$ goes from $0$ to $\sqrt2 -1 $, $(s,c)$ forms a curve located in $[0,\frac{1}{\sqrt2}] \times [\frac{1}{\sqrt2},1]$ starting at $(0,1)$ ending at $(\frac{1}{\sqrt2},\frac{1}{\sqrt2})$ while $(c,s)$ forms a curve in $[\frac{1}{\sqrt2},1] \times[0,\frac{1}{\sqrt2}] $ from $(\frac{1}{\sqrt2},\frac{1}{\sqrt2})$ to $(1,0)$.

Let (we'll need them later to write the distribution of $V$, exactly 8 "mirrored" parts)

$\Gamma_{\pm1,\pm1,1} = \{ t \in \mathbb R^2: t=(\pm\frac{2x}{1+x^2},\pm\frac{1-x^2}{1+x^2}), x\in[0,\sqrt2-1]\}$, $\Gamma_{\pm1,\pm1,-1} = \{ t \in \mathbb R^2: t=(\pm\frac{1-x^2}{1+x^2},\pm\frac{2x}{1+x^2}), x\in[0,\sqrt2-1]\}$

$\Gamma = \bigcup (\Gamma_{\pm1,\pm1,1} \cup \Gamma_{\pm1,\pm1,-1})$ ( glue all these eight pieces together)

Note: $\Gamma$ is just an unit circle! (Adding squared values of coordinates gives us $1$)

So, we should be able to "transform" those coordinates into $\sin, \cos$.

Note that, when $X=\tan(\frac{\alpha}{2})$, then $V=(W,U)=(\sin(\alpha),\cos(\alpha))$

And the density of $X$ tell us that the $\frac{\alpha}{2}$ is uniformly distributed on $[0,\frac{\pi}{8}]$ (taking this $\arctan(\sqrt2 - 1)$) So $\alpha \sim \mathcal U([0,\frac{\pi}{4}])$. But we have eight connected pieces curves $\Gamma_{\pm1,\pm1,\pm1}$, that behave as whole unit circle (those $\sin,\cos$ swaps with themselves and swaps signs).

That means, after letting $S=\{ (x,y) \in \mathbb R^2 : x^2 + y^2 = 1 \} $, the distribution of vector $V$ is $\mu_{_V}$, where for every borel set $A \in \mathcal B(\mathbb R^2) $, we have:

$\mu_{_V}(A) = \frac{1}{8}\int_{A \cap S} \frac{4}{\pi} d\sigma_2(x,y) = \frac{1}{2\pi}\int_{A \cap S}d\sigma_2(x,y)$

To clarify: That $\frac{1}{8}$ is due to our eight pieces with are really similar and mirrored (that is all those probabilities were $\frac{1}{8}$) and the $\frac{4}{\pi}$ is just the density function of $\alpha$ (note that there is indicator function, but in the limits (that $\Gamma$ ).)

After thinking a while, it wasn't that important to find densities of $S,C$, but I didn't see that vector $V$ "forms" a circle before having those calculated.


OK, let's go by steps.

First, for $T$, there is some rejection sampling happening. Noticing that and recalling the Cauchy distribution, we see that $T$ is drawn from a truncated Cauchy distribution -- i.e. $T$ has the distribution of a standard Cauchy random variable, conditioned on lying in the interval $[0,\sqrt{2}-1]$.

What that ends up meaning is that $T$ has density $f_T$ equal to 0 outside of $[0,\sqrt{2}-1]$ and otherwise equal to \[ f_T(x)=\frac{1}{Z} \frac{1}{1 + x^2} \] where \[ Z = \int_0^{\sqrt{2} - 1} \frac{1}{1 + x^2} \, dx = \frac{\pi}{8} \] thankfully -- someone has been nice with the constants. So $f_T(x)=\frac{8}{\pi}\frac{1}{1+x^2}$.

Now for $S$. What is this thing? If $g(x)=2\frac{x}{1+x^2}$ then $S=g(T)$. $g$ is a smooth invertible (in fact strictly increasing) function on the domain we care about (the values that $T$ can take, which are $[0,\sqrt{2}-1]$), so we can use a change of variables formula to compute the density $f_S$ of $S$ as follows.

First, since $g(0)=0$ and $g(\sqrt{2}-1)=\sqrt{2}^{-1}$ (again we have been blessed by someone here), we see that the probability that $S$ lies outside the interval $[0,\sqrt{2}^{-1}]$ is 0, so that $f_S$ is equal to 0 outside that interval, and inside of it we have\[ f_S(x)=f_T(g^{-1}(x)) g^{-1}{'}(x). \] On our region of interest, $g^{-1}(x)=\frac{1-\sqrt{1-x^2}}{x}$, which has derivative\[ g^{-1}{'}(x)=\frac{1}{x^2\sqrt{1-x^2}} - \frac{1}{x^2}. \] OK, so it's a bit of a pain, but \[ f_S(x)=\frac{8}{\pi}\frac{x^2}{(\sqrt{1-x^2}-x-1)^2}\bigg(\frac{1}{x^2\sqrt{1-x^2}} - \frac{1}{x^2}\bigg)=\frac{4}{\pi}\frac{1}{(x+1)\sqrt{1-x^2}}, \] which isn't so bad in the end.

OK, so we have now the distribution of $S$. Let's find that of $C$, noticing that if $h(x)=1-g(x)x$, then $C=h(T)$. But wait -- simplifying $h$ we see that $h(x)=\frac{1}{1+x^2}$. (Who made this problem, and why? Haha.) On the interval we care about,\[ h^{-1}(x)=\frac{\sqrt{1-x}}{\sqrt{x}}, \] with derivative \[h^{-1}{'}(x)=-\frac{1}{2 x \sqrt{x}\sqrt{1-x}},\] so that\[ f_C(x)=f_T(h^{-1}(x)) h^{-1}{'}(x) = ... \] OK, I have to go offline for a while, but this might get you started...


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