For $0\le k\le n$, the adjacent binomial coefficients obeySince $\left(\genfrac{}{}{0ex}{}{n}{k}\right)>0$ and $k+1>0$, the sign is determined by the numerator:The printed codomain ${\mathbb{N}}_{+}$ is inconsistent with the displayed map: $g(n)=-1$, and when $n$ is odd, $g((n-1)/2)=0$. With the usual extension $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=0$ for $k>n$, $g$ is a function ${\mathbb{N}}_0\to \mathbb{Z}$, and $g(k)=0$ for $k>n$; the positive range above remains unchanged. Without that extension, the natural domain is $\{0,\dots ,n\}$.^[ref]

Apply the factorial formula for the binomial coefficients:For $n=0$ or $n=1$, the same equality holds with the conventional zero value when the lower index exceeds the upper index.

There are $n(n-1)\cdots (n-k+1)$ ordered selections without replacement. Each unordered $k$-element subset occurs in exactly $k!$ orders, so the binomial coefficient isOne precise specification of its domain, codomain, and map isHere ${\mathbb{N}}_0=\{0,1,2,\dots \}$, and $0!=1$ includes the single empty selection. The actual values on $D$ are positive, so ${\mathbb{N}}_{+}$ is also a valid codomain. For fixed $n$, one can instead use the function $C_n:\{0,\dots ,n\}\to {\mathbb{N}}_{+}$ with $C_n(k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$.

Try a linear function $f(x)=cx$. The functional equation becomes $c^{2}x=(c+1)x$ for all $x$, so $c$ must be a root of a polynomial $c^2-c-1$. Its two roots of a polynomial giveFor either choice, $c^2=c+1$ proves the required equality directly. The positive coefficient is the golden ratio.

Seek an affine function $f(x)=cx+d$. Its function composition with itself isEquating coefficients requires $c^2=a$ and $(c+1)d=b$. Because $a>0$ and $a\ne 1$, either choice $c=\sqrt{a}$ or $c=-\sqrt{a}$ has $c+1\ne 0$. Thus two valid functions areSubstitution into $c^{2}x+(c+1)d$ verifies each answer. These provide the requested maps; no assumption that every possible solution is affine function is needed.

Take the functionThen $f(f(x))=-(-x)=x$ for every real number $x$, while $f(1)=-1\ne 1$. Thus it is an involution different from the identity function.

If $f(x)=f(y)$, applying the function $f$ gives $f(f(x))=f(f(y))$. Since $f\circ f={\mathrm{id}}_A$, this implies $x=y$, proving injectivity. Also every $y\in A$ is the image of $f(y)$, because $f(f(y))=y$, proving surjectivity.

Thus $f$ is a bijection and its inverse function is itself:Such a function is an involution.

The identity function fixes every element, and function composition means applying the inner function first:

We use a double inclusion proof. First, let $x\in \overline{A\oplus B}$. If $x\in A$, then $x\in B$, since otherwise $x\in A\setminus B\subseteq A\oplus B$. Hence in this case $x\in A\cap B$. If $x\notin A$, then $x\notin B$, since otherwise $x\in B\setminus A\subseteq A\oplus B$. Hence in this case $x\in \overline{A\cup B}$. This proves the first set inclusion.

Conversely, an element of $A\cap B$ belongs to neither $A\setminus B$ nor $B\setminus A$. An element of $\overline{A\cup B}$ also belongs to neither set difference. Thus every element of $(A\cap B)\cup \overline{A\cup B}$ lies outside $A\oplus B$, proving the reverse set inclusion. Therefore

Using set difference as an intersection with a complement of a set, and using double negation, givesInterchanging the two sets simply interchanges the two terms of this union, whose commutativity gives

Every complement of a set is taken in the same universal set $\mathcal{U}$. Apply De Morgan's laws for propositional logic to the membership propositions. For every $x\in \mathcal{U}$,Equality of membership for every element proves the set identities

The set difference consists of elements belonging to the first set but not the second:The symmetric difference consists of elements belonging to exactly one of the two sets:In particular, the two sets in this union are disjoint sets.

The standard normal distribution has mean zero and variance one, so $\mathbb{E}[Z^2]=\mathrm{var}(Z)+(\mathbb{E}[Z]{)}^2=1$. Part (iv) therefore givesUsing the second moment from part (iii),

By the density formula for expected value,For the standard normal distribution, the integrand in the second moment is even, soComparing these two integrals gives

Since $X\ge 0$, the function $X\mapsto X^2$ is strictly increasing on its support. For $y\ge 0$,and $F_Y(y)=0$ for $y<0$. Differentiating the cumulative distribution function givesThe density value at the single point $y=0$ is immaterial. Hence $Y$ has an exponential distribution with rate $1/2$. Direct integration by parts, or its standard expected value, yields

The cumulative distribution function is continuous and strictly increasing for $x>0$, so the median is the unique positive $c$ satisfying $F_X(c)=1/2$. Consequently $e^{-c^2/2}=1/2$, givingThere is no atom at $c$, so $\mathbb{P}(X>c)=1-F_X(c)=1/2$ as required.

Since $X$ is nonnegative, $F_X(x)=0$ for $x<0$. For $x\ge 0$, use $\frac{d}{dt}{e}^{-t^2/2}=-t{e}^{-t^2/2}$ to obtainThus the cumulative distribution function isIn particular it tends to one as $x\to \infty$, confirming normalization of this unit-scale Rayleigh distribution.

We need $0<\alpha <2$. Indeed, if $\alpha <0$, the function is negative on $(\alpha ,0)$; if $\alpha =0$, the improper integral diverges at zero; and if $\alpha \ge 2$, its support interval is empty. For $0<\alpha <2$,Setting this equal to one gives the unique admissible valueOn this interval $1/x>0$, so both requirements for a probability density function are met.

The cosine function is positive on $(0,\pi /2)$ and negative on $(\pi /2,2)$. Thus any nonzero $\alpha$ makes $\alpha \cos x$ negative on an interval of positive length. Taking $\alpha =0$ makes the total definite integral zero. HenceNormalization alone would suggest $\alpha =1/\sin 2$, but that candidate still violates nonnegativity and therefore is not a probability density function.

A probability density function must be nonnegative and have definite integral one. HereThereforeThis value is positive, so the nonnegativity requirement also holds.