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Iosif Pinelis
Iosif Pinelis
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For $t>0$, we have $v^2+bt\le\max(2v^2,2bt)$$v^2+bt\le\max(2bt,2v^2)$, whence \begin{equation} P(X\ge t)\le1\wedge\max[Q_1(t),Q_2(t)]\le[1\wedge Q_1(t)]+[1\wedge Q_2(t)], \end{equation} where $x\wedge y:=\min(x,y)$, \begin{equation} Q_1(t):=C\exp\Big(-\frac{t^2}{4v^2}\Big),\quad Q_2(t):=C\exp\Big(-\frac{t^2}{4bt}\Big). \end{equation} Next, for $t>0$ we have \begin{equation} 1\le Q_1(t)\iff t\le t_1:=2v\sqrt{\ln C}, \end{equation} \begin{equation} 1\le Q_2(t)\iff t\le t_2:=4b\ln C. \end{equation} So\begin{equation} Q_1(t):=C\exp\Big(-\frac{t^2}{4bt}\Big),\quad Q_2(t):=C\exp\Big(-\frac{t^2}{4v^2}\Big). \end{equation} So, \begin{equation} EX=\int_0^\infty P(X\ge t)\,dt\le I_1+I_2, \end{equation} where \begin{multline} I_1:=\int_0^\infty[1\wedge Q_1(t)]\,dt=t_1+\int_{t_1}^\infty Q_1(t)\,dt \\ =t_1+Cv\sqrt 2\int_{\sqrt{2\ln C}}^\infty e^{-s^2/2}\,ds \le t_1+v\sqrt\pi =2v\big(\sqrt{\ln C}+\sqrt\pi/2\big) \end{multline}\begin{multline} I_1:=\int_0^\infty[1\wedge Q_1(t)]\,dt =4b\ln C+C\int_{4b\ln C}^\infty \exp\Big(-\frac{t}{4b}\Big)\,dt=4b(1+\ln C) \end{multline} and \begin{multline} I_2:=\int_0^\infty[1\wedge Q_2(t)]\,dt=t_2+\int_{t_2}^\infty Q_2(t)\,dt \\ =t_2+C\int_{t_2}^\infty \exp\Big(-\frac{t}{4b}\Big)=t_2+4b=4b(1+\ln C); \end{multline}\begin{multline} I_2:=\int_0^\infty[1\wedge Q_2(t)]\,dt =v\sqrt 2\int_0^\infty\min(1,Ce^{-s^2/2})\,ds \\ =v\sqrt 2\Big(\sqrt{2\ln C}+C\int_{\sqrt{2\ln C}}^\infty e^{-s^2/2}\,ds\Big) \le 2v\big(\sqrt{\ln C}+\sqrt\pi/2\big); \end{multline} the latter inequality in the penultimate display follows because $\int_t^\infty e^{-s^2/2}\,ds\le\sqrt{\pi/2}\,e^{-t^2/2}$ for $t\ge0$.

Thus, \begin{equation} EX\le 2v\big(\sqrt{\ln C}+\sqrt\pi/2\big)+4b(1+\ln C), \end{equation}\begin{equation} EX\le 4b(1+\ln C)+2v\big(\sqrt{\ln C}+\sqrt\pi/2\big), \end{equation} which is a bit better than the bound requested in the OP, where we see $\sqrt\pi$ in place of $\sqrt\pi/2$.

For $t>0$, we have $v^2+bt\le\max(2v^2,2bt)$, whence \begin{equation} P(X\ge t)\le1\wedge\max[Q_1(t),Q_2(t)]\le[1\wedge Q_1(t)]+[1\wedge Q_2(t)], \end{equation} where $x\wedge y:=\min(x,y)$, \begin{equation} Q_1(t):=C\exp\Big(-\frac{t^2}{4v^2}\Big),\quad Q_2(t):=C\exp\Big(-\frac{t^2}{4bt}\Big). \end{equation} Next, for $t>0$ we have \begin{equation} 1\le Q_1(t)\iff t\le t_1:=2v\sqrt{\ln C}, \end{equation} \begin{equation} 1\le Q_2(t)\iff t\le t_2:=4b\ln C. \end{equation} So, \begin{equation} EX=\int_0^\infty P(X\ge t)\,dt\le I_1+I_2, \end{equation} where \begin{multline} I_1:=\int_0^\infty[1\wedge Q_1(t)]\,dt=t_1+\int_{t_1}^\infty Q_1(t)\,dt \\ =t_1+Cv\sqrt 2\int_{\sqrt{2\ln C}}^\infty e^{-s^2/2}\,ds \le t_1+v\sqrt\pi =2v\big(\sqrt{\ln C}+\sqrt\pi/2\big) \end{multline} and \begin{multline} I_2:=\int_0^\infty[1\wedge Q_2(t)]\,dt=t_2+\int_{t_2}^\infty Q_2(t)\,dt \\ =t_2+C\int_{t_2}^\infty \exp\Big(-\frac{t}{4b}\Big)=t_2+4b=4b(1+\ln C); \end{multline} the inequality in the penultimate display follows because $\int_t^\infty e^{-s^2/2}\,ds\le\sqrt{\pi/2}\,e^{-t^2/2}$ for $t\ge0$.

Thus, \begin{equation} EX\le 2v\big(\sqrt{\ln C}+\sqrt\pi/2\big)+4b(1+\ln C), \end{equation} which is a bit better than the bound requested in the OP, where we see $\sqrt\pi$ in place of $\sqrt\pi/2$.

For $t>0$, we have $v^2+bt\le\max(2bt,2v^2)$, whence \begin{equation} P(X\ge t)\le1\wedge\max[Q_1(t),Q_2(t)]\le[1\wedge Q_1(t)]+[1\wedge Q_2(t)], \end{equation} where $x\wedge y:=\min(x,y)$, \begin{equation} Q_1(t):=C\exp\Big(-\frac{t^2}{4bt}\Big),\quad Q_2(t):=C\exp\Big(-\frac{t^2}{4v^2}\Big). \end{equation} So, \begin{equation} EX=\int_0^\infty P(X\ge t)\,dt\le I_1+I_2, \end{equation} where \begin{multline} I_1:=\int_0^\infty[1\wedge Q_1(t)]\,dt =4b\ln C+C\int_{4b\ln C}^\infty \exp\Big(-\frac{t}{4b}\Big)\,dt=4b(1+\ln C) \end{multline} and \begin{multline} I_2:=\int_0^\infty[1\wedge Q_2(t)]\,dt =v\sqrt 2\int_0^\infty\min(1,Ce^{-s^2/2})\,ds \\ =v\sqrt 2\Big(\sqrt{2\ln C}+C\int_{\sqrt{2\ln C}}^\infty e^{-s^2/2}\,ds\Big) \le 2v\big(\sqrt{\ln C}+\sqrt\pi/2\big); \end{multline} the latter inequality follows because $\int_t^\infty e^{-s^2/2}\,ds\le\sqrt{\pi/2}\,e^{-t^2/2}$ for $t\ge0$.

Thus, \begin{equation} EX\le 4b(1+\ln C)+2v\big(\sqrt{\ln C}+\sqrt\pi/2\big), \end{equation} which is a bit better than the bound requested in the OP, where we see $\sqrt\pi$ in place of $\sqrt\pi/2$.

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Iosif Pinelis
Iosif Pinelis
  • 127.7k
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For $t>0$, we have $v^2+bt\le\max(2v^2,2bt)$, whence \begin{equation} P(X\ge t)\le \max[1\wedge Q_1(t),1\wedge Q_2(t)]\le[1\wedge Q_1(t)]+[1\wedge Q_2(t)], \end{equation}\begin{equation} P(X\ge t)\le1\wedge\max[Q_1(t),Q_2(t)]\le[1\wedge Q_1(t)]+[1\wedge Q_2(t)], \end{equation} where $x\wedge y:=\min(x,y)$, \begin{equation} Q_1(t):=C\exp\Big(-\frac{t^2}{4v^2}\Big),\quad Q_2(t):=C\exp\Big(-\frac{t^2}{4bt}\Big). \end{equation} Next, for $t>0$ we have \begin{equation} 1\le Q_1(t)\iff t\le t_1:=2v\sqrt{\ln C}, \end{equation} \begin{equation} 1\le Q_2(t)\iff t\le t_2:=4b\ln C. \end{equation} So, \begin{equation} EX=\int_0^\infty P(X\ge t)\,dt\le I_1+I_2, \end{equation} where \begin{multline} I_1:=\int_0^\infty[1\wedge Q_1(t)]\,dt=t_1+\int_{t_1}^\infty Q_1(t)\,dt \\ =t_1+Cv\sqrt 2\int_{\sqrt{2\ln C}}^\infty e^{-s^2/2}\,ds \le t_1+v\sqrt\pi =2v\big(\sqrt{\ln C}+\sqrt\pi/2\big) \end{multline} and \begin{multline} I_2:=\int_0^\infty[1\wedge Q_2(t)]\,dt=t_2+\int_{t_2}^\infty Q_2(t)\,dt \\ =t_2+C\int_{t_2}^\infty \exp\Big(-\frac{t}{4b}\Big)=t_2+4b=4b(1+\ln C); \end{multline} the inequality in the penultimate display follows because $\int_t^\infty e^{-s^2/2}\,ds\le\sqrt{\pi/2}\,e^{-t^2/2}$ for $t\ge0$.

Thus, \begin{equation} EX\le 2v\big(\sqrt{\ln C}+\sqrt\pi/2\big)+4b(1+\ln C), \end{equation} which is a bit better than the bound requested in the OP, where we see $\sqrt\pi$ in place of $\sqrt\pi/2$.

For $t>0$, we have $v^2+bt\le\max(2v^2,2bt)$, whence \begin{equation} P(X\ge t)\le \max[1\wedge Q_1(t),1\wedge Q_2(t)]\le[1\wedge Q_1(t)]+[1\wedge Q_2(t)], \end{equation} where $x\wedge y:=\min(x,y)$, \begin{equation} Q_1(t):=C\exp\Big(-\frac{t^2}{4v^2}\Big),\quad Q_2(t):=C\exp\Big(-\frac{t^2}{4bt}\Big). \end{equation} Next, for $t>0$ we have \begin{equation} 1\le Q_1(t)\iff t\le t_1:=2v\sqrt{\ln C}, \end{equation} \begin{equation} 1\le Q_2(t)\iff t\le t_2:=4b\ln C. \end{equation} So, \begin{equation} EX=\int_0^\infty P(X\ge t)\,dt\le I_1+I_2, \end{equation} where \begin{multline} I_1:=\int_0^\infty[1\wedge Q_1(t)]\,dt=t_1+\int_{t_1}^\infty Q_1(t)\,dt \\ =t_1+Cv\sqrt 2\int_{\sqrt{2\ln C}}^\infty e^{-s^2/2}\,ds \le t_1+v\sqrt\pi =2v\big(\sqrt{\ln C}+\sqrt\pi/2\big) \end{multline} and \begin{multline} I_2:=\int_0^\infty[1\wedge Q_2(t)]\,dt=t_2+\int_{t_2}^\infty Q_2(t)\,dt \\ =t_2+C\int_{t_2}^\infty \exp\Big(-\frac{t}{4b}\Big)=t_2+4b=4b(1+\ln C); \end{multline} the inequality in the penultimate display follows because $\int_t^\infty e^{-s^2/2}\,ds\le\sqrt{\pi/2}\,e^{-t^2/2}$ for $t\ge0$.

Thus, \begin{equation} EX\le 2v\big(\sqrt{\ln C}+\sqrt\pi/2\big)+4b(1+\ln C), \end{equation} which is a bit better than the bound requested in the OP, where we see $\sqrt\pi$ in place of $\sqrt\pi/2$.

For $t>0$, we have $v^2+bt\le\max(2v^2,2bt)$, whence \begin{equation} P(X\ge t)\le1\wedge\max[Q_1(t),Q_2(t)]\le[1\wedge Q_1(t)]+[1\wedge Q_2(t)], \end{equation} where $x\wedge y:=\min(x,y)$, \begin{equation} Q_1(t):=C\exp\Big(-\frac{t^2}{4v^2}\Big),\quad Q_2(t):=C\exp\Big(-\frac{t^2}{4bt}\Big). \end{equation} Next, for $t>0$ we have \begin{equation} 1\le Q_1(t)\iff t\le t_1:=2v\sqrt{\ln C}, \end{equation} \begin{equation} 1\le Q_2(t)\iff t\le t_2:=4b\ln C. \end{equation} So, \begin{equation} EX=\int_0^\infty P(X\ge t)\,dt\le I_1+I_2, \end{equation} where \begin{multline} I_1:=\int_0^\infty[1\wedge Q_1(t)]\,dt=t_1+\int_{t_1}^\infty Q_1(t)\,dt \\ =t_1+Cv\sqrt 2\int_{\sqrt{2\ln C}}^\infty e^{-s^2/2}\,ds \le t_1+v\sqrt\pi =2v\big(\sqrt{\ln C}+\sqrt\pi/2\big) \end{multline} and \begin{multline} I_2:=\int_0^\infty[1\wedge Q_2(t)]\,dt=t_2+\int_{t_2}^\infty Q_2(t)\,dt \\ =t_2+C\int_{t_2}^\infty \exp\Big(-\frac{t}{4b}\Big)=t_2+4b=4b(1+\ln C); \end{multline} the inequality in the penultimate display follows because $\int_t^\infty e^{-s^2/2}\,ds\le\sqrt{\pi/2}\,e^{-t^2/2}$ for $t\ge0$.

Thus, \begin{equation} EX\le 2v\big(\sqrt{\ln C}+\sqrt\pi/2\big)+4b(1+\ln C), \end{equation} which is a bit better than the bound requested in the OP, where we see $\sqrt\pi$ in place of $\sqrt\pi/2$.

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Iosif Pinelis
Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229

For $t>0$, we have $v^2+bt\le\max(2v^2,2bt)$, whence \begin{equation} P(X\ge t)\le \max[1\wedge Q_1(t),1\wedge Q_2(t)]\le[1\wedge Q_1(t)]+[1\wedge Q_2(t)], \end{equation} where $x\wedge y:=\min(x,y)$, \begin{equation} Q_1(t):=C\exp\Big(-\frac{t^2}{4v^2}\Big),\quad Q_2(t):=C\exp\Big(-\frac{t^2}{4bt}\Big). \end{equation} Next, for $t>0$ we have \begin{equation} 1\le Q_1(t)\iff t\le t_1:=2v\sqrt{\ln C}, \end{equation} \begin{equation} 1\le Q_2(t)\iff t\le t_2:=4b\ln C. \end{equation} So, \begin{equation} EX=\int_0^\infty P(X\ge t)\,dt\le I_1+I_2, \end{equation} where \begin{multline} I_1:=\int_0^\infty[1\wedge Q_1(t)]\,dt=t_1+\int_{t_1}^\infty Q_1(t)\,dt \\ =t_1+Cv\sqrt 2\int_{\sqrt{2\ln C}}^\infty e^{-s^2/2}\,ds \le t_1+v\sqrt\pi =2v\big(\sqrt{\ln C}+\sqrt\pi/2\big) \end{multline} and \begin{multline} I_2:=\int_0^\infty[1\wedge Q_2(t)]\,dt=t_2+\int_{t_2}^\infty Q_2(t)\,dt \\ =t_2+C\int_{t_2}^\infty \exp\Big(-\frac{t}{4b}\Big)=t_2+4b=4b(1+\ln C); \end{multline} the inequality in the penultimate display follows because $\int_t^\infty e^{-s^2/2}\,ds\le\sqrt{\pi/2}\,e^{-t^2/2}$ for $t\ge0$.

Thus, \begin{equation} EX\le 2v\big(\sqrt{\ln C}+\sqrt\pi/2\big)+4b(1+\ln C), \end{equation} which is a bit better than the bound requested in the OP, where we see $\sqrt\pi$ in place of $\sqrt\pi/2$.