Isomorphic Spaces

As in Section 1, the standard positive semigroup \(([0, \infty), +)\) has the usual Borel \(\sigma\)-algebra, and has Lebesgue measure \(\lambda\) as the invariant measure. Now let \(S\) be an interval of real numbers of the form \([a, b)\) where \(-\infty \lt a \lt b \leq \infty\) or of the form \((a, b]\) where \(-\infty \leq a \lt b \lt \infty\). The \(\sigma\)-algebra \(\ms S\) is the Borel \(\sigma\)-algebra of subsets of \(S\). Let \(\varphi\) be a continuous, one-to-one function from \(S\) onto \([0, \infty)\).

Define the operator \(\oplus\) on \(S\) by \[x \oplus y = \varphi^{-1}[\varphi(x) + \varphi(y)], \quad x, \, y \in S\] and define the measure \(\mu\) on \((S, \ms S)\) by \[\mu(A) = \lambda[\varphi(A)], \quad A \in \ms S\] With \((S, \ms S, \mu)\) as the underlying measure space, the positive semigroup \((S, \oplus)\) is isomorphic to \(([0, \infty), +)\), and \(\varphi\) is an isomorphism. The endpoint of \(S\) is the identity. The order \(\preceq\) associated with \((S, \oplus)\) is the ordinary order \(\le\) if \(S = [a, b)\) and the reverse \(\ge\) if \(S = (a, b]\).

Details:

From the assumptions on \(\varphi\), the inverse function \(\varphi^{-1}\) is also continuous and maps \([0, \infty)\) one-to-one onto \(S\), so that \(\varphi\) is a topological homeomorphism. The operation \(\oplus\) makes sense and is measurable. In particular, \[\varphi(x \oplus y) = \varphi(x) + \varphi(y), \quad x, \, y \in S\] Note that \(\varphi\) is strictly increasing if \(S = [a, b)\) with \(\varphi(a) = 0\), and \(\varphi\) is strictly decreasing if \(S = (a, b]\) with \(\varphi(b) = 0\).

We use \(\oplus\) for the operator to emphasize the isomorphism with the standard space \(([0, \infty), +)\) and in particular because the operator is commutative. If \(x, y \in S\) and \(x \preceq y\) then \[y \ominus x = \varphi^{-1}[\varphi(y) - \varphi(x)]\] replacing our generic notation \(x^{-1} y\). We can extend the isomorphism to products by the rule \(c \odot x = \varphi^{-1}[c \varphi(x)]\) for \(x \in S\) and \(c \in [0, \infty)\). This generalizes our generic notation \(x^n\) for \(x \in S\) and \(n \in \N\). The measure \(\mu\) is invariant for \((S, \oplus)\) and is unique up to multiplcation by positive constants. In fact, \(\mu\) is the Lebesgue-Stieltjes measure associated with \(\varphi\). If \(\varphi\) is differentiabale with then \[d\mu(x) = |\varphi^\prime(x)| \, dx\] That is, \(\mu\) is absolutely continuous with respect to \(\lambda\) with density \(|\varphi^\prime|\).

For the graph \( (S, \preceq, \mu) \)

The path function \(\gamma_n\) of order \(n \in \N\) is given by \[\gamma_n(x) = \frac{\varphi^n(x)}{n!}, \quad x \in S\]
The path generating function \(\Gamma\) is given by \[\Gamma(x, t) = \exp[t \varphi(x)], \quad x \in S, \, t \in \R \]

Suppose now that \(X\) is a random variable in \(S\) so that \(Y = \varphi(X)\) is a random variable in \([0, \infty)\). Because the spaces are isomorphic, all of the basic functions and concepts for \(X\) relative to \((S, \oplus, \mu)\) are equivalent to those of \(Y\) relative to \(([0, \infty), +, \lambda)\). The following proposition gives a summary. For clairty, we will explicitly give the structures that are necessary.

Suppose that \(X\) is a random variable in \(S\) and that \(Y = \varphi(X)\).

If \(Y\) has reliability function \(F\) for \(([0, \infty), \le)\) then \(X \) has reliability function \(F \circ \varphi\) for \((S, \preceq)\).
If \(Y\) has density function \(f\) with respect to \(\lambda\) then \(X\) has density function \(f \circ \varphi\) with respect to \(\mu\).
If \(Y\) has rate function \(r\) for \(([0, \infty), \le, \lambda)\) then \(X\) has rate function \(r \circ \varphi\) for \((S, \preceq, \mu)\).
If \(Y\) has cumulative rate function \(R\) for \(([0, \infty), \le)\) then \(X\) has cumulative rate function \(R \circ \varphi\) for \((S, \preceq)\).
If \(Y\) has average rate function \(\bar r\) for \(([0, \infty), \le, \lambda)\) then \(X\) has average rate function \(\bar r \circ \varphi\) for \((S, \preceq, \lambda)\).

Ordinarily, part (d) requires a reference to the underlying measures, but in this setting we can avoid that since the cumulative rate function of \(Y\) for \(([0, \infty), \le)\) is \(R = - \ln F\) and similarly, the cumulative rate function of \(X\) for \((S, \preceq)\) is \(R \circ \varphi = - \ln F \circ \varphi\). In the context of (b), if \(\varphi\) is differentiable then \(X\) has density \((f \circ \varphi) |\varphi^\prime|\) with respect to Lebesgue measure \(\lambda\) on \((S, \ms S)\). The following result gives the equivalence between various aging (improvement) properties: increasing (decreasing) failure rate, increasing (decreasing) failure rate average, new better (worse) than used, and the ageless exponential property.

Suppose again that \(X\) is a random variable in \(S\) and that \(Y = \varphi(X)\).

\(X\) is IFR (DFR) on \((S, \preceq)\) if and only if \(Y\) is IFR (DFR) on \(([0, \infty), \le)\).
\(X\) is IFRA (DFRA) on \((S, \preceq, \mu)\) if and only if \(Y\) is IFRA (DFRA) on \(([0, \infty), \le, \lambda)\)
\(X\) is NBU (NWU) on \((S, \oplus)\) if and only if \(Y\) is NBU (NWU) on \(([0, \infty), +)\).
\(X\) has an exponential distribution on \((S, \oplus)\) if and only if \(Y\) has an exponential distribution on \(([0, \infty), +)\).

Note that if \(S = (a, b]\) so that \(\preceq\) is the order \(\ge\), the terms increasing and decreasing in parts (a) and (b) actually mean decreasing and increasing, respectively, in terms of the ordinary order. From part (d), we have the following results:

Suppose again that \(X\) is a random variable in \(S\)

\(X\) is exponential for \((S, \oplus)\) if and only if \(X\) is memoryless for \((S, \oplus)\), if and only if \(X\) has constant rate for \((S, \preceq, \mu)\).
The exponential distribution for \((S, \oplus)\) that has constant rate \(\alpha \in (0, \infty)\) has reliability function \(F\) defined by \[F(x) = \exp[-\alpha \varphi(x)], \quad x \in S\]
The probability density function \(f\) of \(X\) relative to \(\mu\) is given by \[f(x) = \alpha \exp[-\alpha \varphi(x)], \quad x \in S\]
If \(\varphi\) is differentiable, the probability density function of \(X\) relative to Lebesgue measure \(\lambda\) is given by \[x \mapsto \alpha \exp[-\alpha \varphi(x)] |\varphi^\prime(x)|, \quad x \in S\]

In particular, note that \(\alpha |\varphi^\prime|\) is the failure rate function of \(X\) for \((S, \le)\). Every continuous distribution on \(S\) is exponential relative to some semigroup.

Suppose that \(X\) is a random variable with a continuous distribution on \(S\). Then \(X\) is exponential with respect to a semigroup \((S, \oplus)\) isomorphic to \(([0, \infty), +)\). Specifically, \(X\) is exponential on the semigroup \((S, \oplus)\), with isomorphism \(\varphi\) defined as follows:

If \(S = [a, b)\), let \(\varphi(x) = -\ln [\P(X \ge x)]\) for \(x \in [a, b)\).
if \(S = (a, b]\), let \(\varphi(x) = -\ln [\P(X \le x)]\) for \(x \in (a, b]\).

This idea will be studied further in Section 3 on relative aging. The following result follows from the discussion of entropy in Section 1.5

Suppose that \(X\) is exponential on \((S, \oplus)\) with rate parameter \(\alpha \in (0, \infty)\) and reliability function \(F\) given above. Then \(X\) maximizes entropy over all random variables \(Y\) in \(S\) with \[\E[\varphi(Y)] = \E[\varphi(X)] = 1 / \alpha\] The maximum entropy is \(1 - \ln \alpha\).

Suppose again that \(X\) is exponential on \((S, \oplus)\) with rate parameter \(\alpha \in (0, \infty)\). Let \(\bs {Y} = (Y_1, Y_2, \ldots)\) is the random walk on \((S, \preceq)\) or equivalently the random walk on \((S, \oplus)\) associated with \(X\). Then \((\varphi(Y_1), \varphi(Y_2), \ldots)\) is the random walk on \(([0, \infty), \le)\) or equivalently the random walk on \(([0, \infty), +)\), respectively, associated with \(\varphi(X)\). In particular,

The density function \(f_n\) of \(Y_n\) with respect \(\mu\) is \[f_n(x) = \alpha^n \frac{\varphi^{n-1}(x)}{(n - 1)!} \exp[-\alpha \varphi(x)], \quad x \in S\]
If \(\varphi\) is differentiable, the density \(g_n\) of \(Y_n\) with respect to Lebesgue measure \(\lambda\) is \[g_n(x) = \alpha^n \frac{\varphi^{n-1}(x)}{(n - 1)!} \exp[-\alpha \varphi(x)] \varphi^\prime(x), \quad x \in S\]

The shifted space

Our first example is a rather trivial modification of the standard space, but still leads to some helpful insights.

Let \(S = [a, \infty)\) where \(a \in (0, \infty)\), and define \(\varphi: [a, \infty) \to [0, \infty)\) by \(\varphi(x) = x - a\) so that \(\varphi\) is a homeomorphism from \([a, \infty)\) onto \([0, \infty)\).

The associated operator is \(x \oplus y = x + y - a\) for \(x, \, y \in [a, \infty)\)
The corresponding order is the ordinary order \(\le\).
The invariant measure is ordinary Lebesgue measure

For the graph \( ([a, \infty), \le) \),

The path function \(\gamma_n\) of order \(n \in \N\) is given by \(\gamma_n(x) = (x - a)^n / n!\) for \(x \in [a, \infty)\).
The path generating function \(\Gamma\) is given by \(\Gamma(x, t) = e^{t (x - a)}\) for \(x \in [a, \infty), \, t \in \R\).

Suppose now that \(X\) is a random variable in \([a, \infty)\). The reliability function \(F\) of \(X\) for \(([a, \infty), \le)\) is just the standard one, so that \(F(x) = \P(X \ge x)\) for \(x \in [a, \infty)\).

Suppose that \(X\) has the exponential distribution on \(([a, \infty), \oplus)\) with constant rate \(\beta \in (0, \infty)\).

The reliability function \(F\) of \(X\) is given by \(F(x) = e^{-\beta(x - a)}\) for \(x \in [a, \infty)\).
The density function \(f\) of \(X\) is given by \(f(x) = \beta e^{-\beta(x - a)}\) for \(x \in [a, \infty)\)

So \(X\) has a shifted exponential distribution with parameters \(a\) and \(\beta\).

Suppose again that \(X\) is exponential for \(([a, \infty), \oplus)\) with rate parameter \(\beta \in (0, \infty)\) and let \(\bs Y = (Y_1, Y_2, \ldots)\) denote the random walk corresponding to \(X\). The density function \(f_n\) of \(Y_n\) is given by \[f_n(x) = \beta^n \frac{(x - a)^{n-1}}{(n - 1)!} e^{-\beta (x - a)}, \quad x \in [a, \infty)\]

Of course \(f_n\) is the shifted version of the standard gamma distribution with parameters \(n\) and \(\beta\). But we can also consider \([a, \infty)\) with the standard addition operator \(+\). So \(([a, \infty), +)\) is a strict positive semigroup, and the associated strict partial order \(\prec\) is given by \(x \prec y\) if and only if \(x + a \le y\). Once again we use Lebesgue measure \(\lambda\) as the invariant reference measure. Now we are outside of the setting of the general theory above, but it's interesting to compare the two spaces.

The path function \(\gamma_n\) of order \(n \in \N\) for \(([a, \infty), \prec)\) is given by \(\gamma_n(x) = 0\) for \(x \lt (n + 1)a\) and \[\gamma_n(x) = \frac{[x - (n + 1)a]^n}{n!}, \quad x \ge (n + 1) a\]

Details:

A combinatorial proof is best. Let \(n \in \N_+\) and note first that \(\gamma_n(x) = 0\) for \(x \lt (n + 1)a\). Suppose that \(x \ge (n + 1) a\) and let \(\bs u = (u_1, u_2, \ldots, u_n)\) satisfy \[(n + 1) a \le u_1 \le u_2 \le \cdots \le u_n \le x\] so that \((u_1, u_2, \ldots, u_n, x)\) is a path of length \(n\) in \(([a, \infty), \le)\) terminating in \(x\). Define \(\bs x = (x_1, x_2, \ldots, x_n)\) by \(x_k = u_k - (n - k + 1) a\) for \(k \in \{1, 2, \ldots, n\}\). Then \[x_1 \prec x_2 \prec \cdots \prec x_n \prec x\] so \((x_1, x_2, \ldots, x_n, x)\) is a path of length \(n\) in \(([a, \infty), \prec)\) terminating in \(x\). Conversely, given such a path \(\bs x\) we can recover the path \(\bs u\) by \(u_k = x_k + (n - k + 1)a\) for \(k \in \{1, 2, \ldots, n\}\). The measure of the set of paths \(\bs x\) is the same as the measure of the set of paths \(\bs u\), which is \([(x - (n + 1) a]^n / n!\).

Suppose now that \(X\) is a random variable in \([a, \infty)\). The reliability function \(F\) of \(X\) for \(([a, \infty), \prec)\) is given by \(F(x) = \P(X \ge x + a)\) for \(x \in [a, \infty)\). Clearly \(F\) does not determine the distribution of \(X\) since \(F\) gives no information about the distribution on the interval \([a, 2 a)\). If \(X\) has density \(f\) then \[F(x) = \int_{x+a}^\infty f(t) dt, \quad x \in [a, \infty)\] and hence \(F^\prime(x) = - f(x + a)\) for \(x \in [a, \infty)\).

Suppose that \(X\) has the shifted exponential distribution on \([a, \infty)\) with parameter \(\beta \in (0, \infty)\). Then \(X\) has an exponential distribution for \(([a, \infty), +)\) with constant rate \(\beta e^{\beta a}\).

Details:

As before, \(X\) has density function \(f\) given by \(f(x) = \beta e^{-\beta(x - a)}\) for \(x \in [a, \infty)\). The reliability function \(F\) of \(X\) for \(([a, \infty), \prec)\) is given by \[F(x) = \P(X \ge x + a) = \P(X - a \ge x) = e^{-\beta x}, \quad x \in [a, \infty)\] So the distribution of \(X\) is memoryless for \(([a, \infty), +)\) and has constant rate \(\beta e^{\beta a}\).

Suppose again that \(X\) has the shifted exponential distribution on \([a, \infty)\) with parameter \(\beta \in (0, \infty)\), and that \(\bs Y = (Y_1, Y_2, \ldots)\) is the random walk on \(([a, \infty), +)\) associated with \(X\). For \(n \in \N_+\), The density function \(f_n\) of \(Y_n\) is given by \[f_n(x) = (\beta e^{\beta a})^n \gamma_{n-1}(x) F(x) = \beta^n \frac{(x - n a)^{n-1}}{(n - 1)!} e^{-\beta (x - n a)}, \quad x \ge n a\]

It's interesting to compare the two spaces. The positive semigroup \(([a, \infty), \oplus)\) corresponds to the standard order \(\le\) but has the non-standard operator \(\oplus\). The strict positive semigroup \(([a, \infty), +)\) has the standard operator \(+\) but has the non-standard order \(\prec\). The shifted exponential distribution on \([a, \infty)\) is an exponential distribution for both spaces, but with different rate parameters.

The Beta Distribution

Let \(S = (0, 1]\) and let \(\varphi(x) = -\ln x\) for \(x \in (0, 1]\) so that \(\varphi\) is a homeomorphism from \((0, 1]\) onto \([0,\infty)\).

The associated operation \(\cdot\) is ordinary multiplication.
The associated order is \(\ge\), the reverse of the ordinary order.
The invariant measure \(\mu\) is given by \(d\mu(x) = (1/x) \, dx\).

For the graph \( ((0, 1], \ge, \mu) \),

The path function \(\gamma_n\) of order \(n \in \N\) is given by \(\gamma_n(x) = (-1)^n \ln^n(x) / n!\) for \(x \in (0, 1]\).
The generating function \(\Gamma\) is given by \( \Gamma(x, t) = 1 / x^t \) for \( x \in (0, 1] \) and \( t \in \R \).

Suppose that \(X\) has the exponential distribution for \( ((0, 1], \cdot) \) with rate parameter \( \alpha \in (0, \infty) \).

The reliability function \(F\) is given by \(F(x) = x^\alpha\) for \(x \in (0, 1]\)
The density \(f\) of \( X \) with respect to the invariant measure \(\mu\) is given by \(f(x) = \alpha x^\alpha\) for \(x \in (0, 1]\).
The density \(g\) of \( X \) with respect to Lebesgue measure \(\lambda\) is given by \(g(x) = \alpha x^{\alpha - 1}\) for \(x \in (0, 1]\)

The distribution of \( X \) is the beta distribution with left parameter \(\alpha\) and right parameter 1, and the special case \(\alpha = 1\) gives the uniform distribution on \((0, 1]\). Random variable \(X\) maximizes entropy over all random variables \(Y\) taking values in \((0, 1]\) with \(\E(-\ln Y) = 1 / \alpha\).

Open the simulation of the beta distribution. The probability density function displayed is the ordinary one, with respect to Lebesgue measure. Vary the rate parameter \(\alpha\) and note the shape of the density function. Run the simulation and compare the empirical density function to the probability density function.

Suppose again that \(X\) has the exponential distribution for \( ((0, 1], \cdot) \) with rate parameter \( \alpha \in (0, \infty)\) and that \(\bs Y = (Y_1, Y_2, \ldots)\) is the random walk associated with \( X \). For \(n \in \N_+\),

The density function \(f_n\) of \(Y_n\) with respect to the invariant measure \(\mu\) is given by \[f_n(x) = \alpha^n (-1)^{n-1} \frac{\ln^{n-1}(x)}{(n - 1)!} x^\alpha, \quad x \in (0, 1]\]
The density \(g_n\) of \(Y_n\) with respect to Lebesgue measure \(\lambda\) is given by \[g_n(x) = \alpha^n (-1)^{n-1} \frac{\ln^{n-1}(x)}{(n - 1)!} x^{\alpha - 1}, \quad x \in (0, 1]\].

The Pareto Distribution

Let \(S = [1, \infty)\) and let \(\varphi(x) = \ln x\) for \(x \in [1, \infty)\) so that \(\varphi\) is a homeomorphism from \([1, \infty)\) onto \([0,\infty)\).

The associated operation is ordinary multiplication \(\cdot\).
The associated order is \(\le\), the ordinary order.
The invariant measure \(\mu\) is given by \(d\mu(x) = (1/x) \, dx\).

For the graph \( ([1, \infty), \le, \mu) \),

The path function \(\gamma_n\) of order \(n \in \N\) is given by \(\gamma_n(x) = \ln^n(x) / n!\) for \(x \in [1, \infty)\).
The generating function \(\Gamma\) is given by \(\Gamma(x, t) = x^t \) for \( x \in [1, \infty) \) and \( t \in \R \).

Suppose that \(X\) has the exponential distribution for \( ([1, \infty), \cdot) \) with rate parameter \( \alpha \in (0, \infty) \).

The reliability function \(F\) of \(X\) is given by \(F(x) = 1 / x^\alpha\) for \(x \in [1, \infty)\)
The density \(f\) of \( X \) with respect to the invariant measure \(\mu\) is given by \(f(x) = \alpha / x^\alpha\) for \(x \in [1, \infty)\).
The density \(g\) of \( X \) with respect to Lebesgue measure \(\lambda\) is given by \(g(x) = \alpha / x^{\alpha + 1}\) for \(x \in [1, \infty)\)

The distribution of \( X \) is the Pareto distribution with shape parameter \(\alpha\). Random variable \(X\) maximizes entropy over all random variables \(Y\) taking values in \([1, \infty)\) with \(\E(\ln Y) = 1 / \alpha\).

Open the simulation of the Pareto distribution. The probability density function displayed is the ordinary one, with respect to Lebesgue measure. Vary the rate parameter \(\alpha\) and note the shape of the density function. Run the simulation and compare the empirical density function to the probability density function.

Suppose again that \(X\) has the exponential distribution for \( ([1, \infty), \cdot) \) with rate parameter \( \alpha \in (0, \infty)\) and that \(\bs Y = (Y_1, Y_2, \ldots)\) is the random walk associated with \( X \). For \(n \in \N_+\),

The density function \(f_n\) of \(Y_n\) with respect to the invariant measure \(\mu\) is given by \[f_n(x) = \alpha^n \frac{\ln^{n - 1}(x)}{(n - 1)!} \frac{1}{x^\alpha}, \quad x \in [1, \infty) \]
The density function \(g_n\) of \(Y_n\) with respect to Lebesgue measure is given by \[g_n(x) = \alpha^n \frac{\ln^{n - 1}(x)}{(n - 1)!} \frac{1}{x^{\alpha + 1}}, \quad x \in [1, \infty) \]

An Application to Brownian Functionals

Let \(S = [0, 1/2)\) and let \(\varphi(x) = x / (1 - 2x)\) for \(x \in [0, 1/2)\) so that \(\varphi\) is a homeomorphism from \([0, 1/2)\) onto \([0,\infty)\).

The associated semigroup operation \(\oplus\) is given by \[x \oplus y = \frac{x + y - 4xy}{1 - 4xy}; \quad x, \, y \in [0, 1/2)\]
The associated order is the ordinary order \(\le\).
The invariant measure \(\mu\) is given by \[d\mu(x) = \frac{1}{(1 - 2x)^2} \, dx\]

The positive semigroup \(([0, 1/2), \oplus)\) occurs in a study of generalized Brownian functionals by Hida.

For the graph \(([0, 1/2), \le, \mu)\),

The path function \(\gamma_n\) of order \(n \in \N\) is given by \[\gamma_n(x) = \frac{1}{n!} \left(\frac{x}{1 - 2x}\right)^n, \quad x \in [0, 1/2)\]
The generating function \(\Gamma\) is given by \[ \Gamma(x, t) = \exp\left(t \frac{x}{1 - 2 x}\right), \quad x \in [0, 1/2), \, t \in \R \]

Suppose that \(X\) has the exponential distribution on \( ([0, 1/2), \oplus) \) with rate parameter \(\alpha \in (0, \infty) \).

The reliability function \(F\) of \(X\) is given by \[F(x) = \exp\left(-\alpha \frac{x}{1 - 2 x}\right), \quad x \in [0, 1/2)\]
The density function \(f\) of \(X\) with respect to the invariant measure \(\mu\) is given by \[f(x) = \alpha \exp\left(-\alpha \frac{x}{1 - 2 x}\right), \quad x \in [0, 1/2)\]
The density function \(g\) of \(X\) with respect to Lebesgue measure \(\lambda\) is given by \[g(x) = \alpha \frac{1}{(1 - 2 x)^2} \exp\left(-\alpha \frac{x}{1 - 2x}\right), \quad x \in [0, 1/2)\]

We will refer to the distribution of \(X\) as the Hida distribution with parameter \(\alpha\). Random variable \(X\) maximizes entropy over all random variables \(Y\) in \([0, 1/2)\) with \(\E[Y / (1 - 2 Y)] = 1 / \alpha\).

Open the simulation of the Hida distribution. The probability density function displayed is the ordinary one, with respect to Lebesgue measure. Vary the rate parameter \(\alpha\) and note the shape of the density function. Run the simulation and compare the empirical density function to the probability density function.

Suppose again that \(X\) has the exponential distribution on \( ([0, 1/2), \oplus) \) with rate parameter \(\alpha \in (0, \infty) \) and that \(\bs Y = (Y_1, Y_2, \ldots)\) is the random walk associated with \(X\). For \(n \in \N_+\),

The density function \(f_n\) of \(Y_n\) with respect to the invariant measure \(\mu\) is given by \[f_n(x) = \frac{\alpha^n}{(n - 1)!} \left(\frac{x}{1 - 2x}\right)^{n - 1} \exp\left(-\alpha \frac{x}{1 - 2 x}\right), \quad x \in [0, 1/2)\]
The density function \(g_n\) of \(Y_n\) with respect to Lebesgue measure \(\lambda\) is given by \[g_n(x) = \frac{\alpha^n}{(n - 1)!} \frac{x^{n - 1}}{(1 - 2 x)^{n + 1}} \exp\left(-\alpha \frac{x}{1 - 2 x}\right), \quad x \in [0, 1/2)\]