Return to Answer

Regarding the initial question, the construction explained by Jeff is my favorite. Regarding the edited version, which asks that $Z_t$ be written as a function of $X$ and $t$, the following construction is ugly but correct.

Assume first that $X$ is uniform on the interval $[0,1]$. One asks that $P(Z_t>z)=(1-z)/(1-t)$ for every $z$ in $[t,1]$ and one knows that $P(X > x)=1-x$ for every $x$ in $[0,1]$. Solving the equation $P(Z_t>z)=P(X>x)=1-x$ for $z$ yields $z=t+(1-t)x$, hence a (pathwise increasing) solution in this specific case is $$ Z_t=t+(1-t)X. $$ In the general case, recall that the complementary cumulative distribution function $G$ of $X$ is defined by $G(x)=P(X>x)$ for every real number $x$. One asks that $G(x)=G(z)/G(t)$, hence a (pathwise nondecreasing) solution in the general case is $$ Z_t=G^{-1}(G(t)G(X)). $$ Here the complementary quantile function $G^{-1}$ of $X$ is defined by the formula $$ G^{-1}(u)=\inf\{x \vert G(x)\le u\}, $$ for (at least) every $u$ in $]0,1[$. As the notation suggests, $G^{-1}$ is an inverse of $G$ in the sense that $G^{-1}(G(X))=X$ almost surely. Equivalently, $G^{-1}(u)\le x$ if and only if $G(x)\le u$.

A nice example is when $X$ is exponential (with any parameter), then $Z_t=t+X$ for every nonnegative $t$. And if $X$ is uniform on $[0,1]$, another solution than the one above is $Z_t=1-(1-t)X$.

Regarding the initial question, the construction explained by Jeff is my favorite. Regarding the edited version, which asks that $Z_t$ be written as a function of $X$ and $t$, the following construction is ugly but correct.

Assume first that $X$ is uniform on the interval $[0,1]$. One asks that $P(Z_t>z)=(1-z)/(1-t)$ for every $z$ in $[t,1]$ and one knows that $P(X > x)=1-x$ for every $x$ in $[0,1]$. Solving the equation $P(Z_t>z)=P(X>x)=1-x$ for $z$ yields $z=t+(1-t)x$, hence a (pathwise increasing) solution in this specific case is $$ Z_t=t+(1-t)X. $$ In the general case, recall that the complementary cumulative distribution function $G$ of $X$ is defined by $G(x)=P(X>x)$ for every real number $x$. One asks that $G(x)=G(z)/G(t)$, hence a (pathwise nondecreasing) solution in the general case is $$ Z_t=G^{-1}(G(t)G(X)). $$ Here the complementary quantile function $G^{-1}$ of $X$ is defined by the formula $$ G^{-1}(u)=\inf\{x \vert G(x)\le u\}, $$ for (at least) every $u$ in $]0,1[$. As the notation suggests, $G^{-1}$ is an inverse of $G$ in the sense that $G^{-1}(G(X))=X$ almost surely. Equivalently, $G^{-1}(u)\le x$ if and only if $G(x)\le u$.

A nice example is when $X$ is exponential (with any parameter), then $Z_t=t+X$ for every nonnegative $t$. A conjugate example is when $X$ follows a power law in the sense that $G(x)=(x_0/x)^a$ for every $x\ge x_0$, for given positive $x_0$ and $a$, then $Z_t=tX/x_0$ for every $t\ge x_0$ and $Z_t=X$ for every $t\le x_0$. And if $X$ is uniform on $[0,1]$, another solution than the one above is $Z_t=1-(1-t)X$.

Regarding the initial question, the construction explained by Jeff is my favorite. Regarding the edited version, which asks that $Z_t$ be written as a function of $X$ and $t$, the following construction is ugly but correct.

Assume first that $X$ is uniform on the interval $[0,1]$. One asks that $P(Z_t>z)=(1-z)/(1-t)$ for every $z$ in $[t,1]$ and one knows that $P(X > x)=1-x$ for every $x$ in $[0,1]$. Solving the equation $P(Z_t>z)=P(X>x)=1-x$ for $z$ yields $z=t+(1-t)x$, hence a (pathwise increasing) solution in this specific case is $$ Z_t=t+(1-t)X. $$ In the general case, recall that the complementary cumulative distribution function $G$ of $X$ is defined by $G(x)=P(X>x)$ for every real number $x$. One asks that $G(x)=G(z)/G(t)$, hence a (pathwise nondecreasing) solution in the general case is $$ Z_t=G^{-1}(G(t)G(X)). $$ Here the complementary quantile function $G^{-1}$ of $X$ is defined by the formula $$ G^{-1}(u)=\inf(x \vert G(x)\le u), $$ for (at least) every $u$ in $]0,1[$. As the notation suggests, $G^{-1}$ is an inverse of $G$ in the sense that $G^{-1}(G(X))=X$ almost surely. Equivalently, $G^{-1}(u)\le x$ if and only if $G(x)\le u$.

A nice example is when $X$ is exponential (with any parameter), then $Z_t=t+X$ for every nonnegative $t$. And if $X$ is uniform on $[0,1]$, another solution than the one above is $Z_t=1-(1-t)X$.

Regarding the initial question, the construction explained by Jeff is my favorite. Regarding the edited version, which asks that $Z_t$ be written as a function of $X$ and $t$, the following construction is ugly but correct.

Assume first that $X$ is uniform on the interval $[0,1]$. One asks that $P(Z_t>z)=(1-z)/(1-t)$ for every $z$ in $[t,1]$ and one knows that $P(X > x)=1-x$ for every $x$ in $[0,1]$. Solving the equation $P(Z_t>z)=P(X>x)=1-x$ for $z$ yields $z=t+(1-t)x$, hence a (pathwise increasing) solution in this specific case is $$ Z_t=t+(1-t)X. $$ In the general case, recall that the complementary cumulative distribution function $G$ of $X$ is defined by $G(x)=P(X>x)$ for every real number $x$. One asks that $G(x)=G(z)/G(t)$, hence a (pathwise nondecreasing) solution in the general case is $$ Z_t=G^{-1}(G(t)G(X)). $$ Here the complementary quantile function $G^{-1}$ of $X$ is defined by the formula $$ G^{-1}(u)=\inf\{x \vert G(x)\le u\}, $$ for (at least) every $u$ in $]0,1[$. As the notation suggests, $G^{-1}$ is an inverse of $G$ in the sense that $G^{-1}(G(X))=X$ almost surely. Equivalently, $G^{-1}(u)\le x$ if and only if $G(x)\le u$.

A nice example is when $X$ is exponential (with any parameter), then $Z_t=t+X$ for every nonnegative $t$. And if $X$ is uniform on $[0,1]$, another solution than the one above is $Z_t=1-(1-t)X$.

Regarding the initial question, the construction explained by Jeff is my favorite. Regarding the edited version, which asks that $Z_t$ be written as a function of $X$ and $t$, the following construction is ugly but correct.

Assume first that $X$ is uniform on the interval $[0,1]$. One asks that $P(Z_t>z)=(1-z)/(1-t)$ for every $z$ in $[t,1]$ and one knows that $P(X > x)=1-x$ for every $x$ in $[0,1]$. Solving the equation $P(Z_t>z)=P(X>x)=1-x$ for $z$ yields $z=t+(1-t)x$, hence a (pathwise increasing) solution in this specific case is $$ Z_t=t+(1-t)X. $$ In the general case, recall that the complementary cumulative distribution function $G$ of $X$ is defined by $G(x)=P(X>x)$ for every real number $x$. One asks that $G(x)=G(z)/G(t)$, hence a (pathwise nondecreasing) solution in the general case is $$ Z_t=G^{-1}(G(t)G(X)). $$ Here the complementary quantile function $G^{-1}$ of $X$ is defined by the formula $$ G^{-1}(u)=\inf(x \vert G(x)\le u), $$ for (at least) every $u$ in $]0,1[$. As the notation suggests, $G^{-1}$ is an inverse of $G$ in the sense that $G^{-1}(u)\le x$ if and only if $G(x)\le u$.

A nice example is when $X$ is exponential (with any parameter), then $Z_t=t+X$ for every nonnegative $t$. And if $X$ is uniform on $[0,1]$, another solution than the one above is $Z_t=1-(1-t)X$.

Regarding the initial question, the construction explained by Jeff is my favorite. Regarding the edited version, which asks that $Z_t$ be written as a function of $X$ and $t$, the following construction is ugly but correct.

Assume first that $X$ is uniform on the interval $[0,1]$. One asks that $P(Z_t>z)=(1-z)/(1-t)$ for every $z$ in $[t,1]$ and one knows that $P(X > x)=1-x$ for every $x$ in $[0,1]$. Solving the equation $P(Z_t>z)=P(X>x)=1-x$ for $z$ yields $z=t+(1-t)x$, hence a (pathwise increasing) solution in this specific case is $$ Z_t=t+(1-t)X. $$ In the general case, recall that the complementary cumulative distribution function $G$ of $X$ is defined by $G(x)=P(X>x)$ for every real number $x$. One asks that $G(x)=G(z)/G(t)$, hence a (pathwise nondecreasing) solution in the general case is $$ Z_t=G^{-1}(G(t)G(X)). $$ Here the complementary quantile function $G^{-1}$ of $X$ is defined by the formula $$ G^{-1}(u)=\inf(x \vert G(x)\le u), $$ for (at least) every $u$ in $]0,1[$. As the notation suggests, $G^{-1}$ is an inverse of $G$ in the sense that $G^{-1}(G(X))=X$ almost surely. Equivalently, $G^{-1}(u)\le x$ if and only if $G(x)\le u$.

A nice example is when $X$ is exponential (with any parameter), then $Z_t=t+X$ for every nonnegative $t$. And if $X$ is uniform on $[0,1]$, another solution than the one above is $Z_t=1-(1-t)X$.