Show that for $x>0$, $c>0$, $P[ X>x+(c/x)\mid X>x]
If $X \sim N(0,\sigma^2)$, show that for $x>0$, $c>0$, $$P\left[ X>x+(c/x)\mid X>x\right]<e^{-c/\sigma^2}.$$
The LHS is equivalent to $$\frac{1-F(x+(c/x))}{1-F(x)} = \frac{\int_{x+(c/x)}^\infty e^{-u^2/(2 \sigma^2)} \, du}{\int_x^\infty e^{-u^2/(2 \sigma^2)} \, du}$$
But I don't know how to proceed then.
 A: Let us stick to elementary tools. You showed yourself that the inequality to prove is $$I(x)\leqslant e^{-c/\sigma^2}\cdot J(x)\tag{1}$$ where $$I(x)=\int_{x+c/x}^\infty e^{-u^2/(2 \sigma^2)} \, du\qquad\text{and}\qquad J(x)=\int_x^\infty e^{-v^2/(2 \sigma^2)} \, dv$$ Surely you will agree that the change of variable $$u=v+c/x\tag{2}$$ applied to $I(x)$ yields exactly the integration interval $v\geqslant x$ of $J(x)$, hence it seems worthwhile to explore at least a little further the effects of the change of variable $(2)$ on $I(x)$, right?
But, in terms of functions to be integrated, note that, for every $v\geqslant x$, $$(v+c/x)^2\geqslant v^2+2vc/x\geqslant v^2+2c$$ hence $$e^{-(v+c/x)^2/(2 \sigma^2)}\leqslant e^{-c/\sigma^2}\cdot e^{-v^2/(2\sigma^2)}\tag{3}$$ Thus, behold! Integrating the pointwise inequality $(3)$ on $v\geqslant x$ yields exactly $(1)$.
A: Note
$$ \frac {1 - \Phi((x + c/x)/\sigma)} {1 - \Phi(x/\sigma)} < e^{-c/\sigma^2} 
\iff g(x) \triangleq 1 - \Phi\left(\frac {x} {\sigma} + \frac {c} {\sigma x}\right) - e^{-c/\sigma^2} + e^{-c/\sigma^2} \Phi\left(\frac {x} {\sigma}\right) < 0$$
Differetiate the $g(x)$ with respect to $x$:
$$ \begin{align} g'(x) 
&= -\frac {1} {\sigma}\phi\left(\frac {x} {\sigma} + \frac {c} {\sigma x}\right)
\left(1- \frac {c} {x^2}\right)
+ \frac {1} {\sigma}e^{-c/\sigma^2} \phi\left(\frac {x} {\sigma}\right) \\
&= e^{-c/\sigma^2} \frac {1} {\sqrt{2\pi\sigma^2}} e^{-x^2/(2\sigma^2)}
- \frac {1} {\sqrt{2\pi\sigma^2}} \left(1- \frac {c} {x^2}\right)
e^{-(x+c/x)^2/(2\sigma^2)} \\
&= \frac {1} {\sqrt{2\pi\sigma^2}}\left[ e^{-c/\sigma^2}e^{-x^2/(2\sigma^2)}
- \left(1- \frac {c} {x^2}\right) e^{(-x^2-2c-c^2/x^2)/(2\sigma^2)}\right] \\
&= \frac { e^{-c/\sigma^2}e^{-x^2/(2\sigma^2)}} {\sqrt{2\pi\sigma^2}}
\left[ 1
- \left(1- \frac {c} {x^2}\right) e^{-c^2/(2\sigma^2x^2)}\right] \\
\end{align}$$
By the assumptions $c > 0$ and $x > 0$, we have
$$ \left(1- \frac {c} {x^2}\right) < 1 \text { and } 0 < e^{-c^2/(2\sigma^2x^2)} < 1$$
and therefore $g'(x) > 0$ for any $x > 0, c > 0$. As a result, $g$ is strictly increasing. To find its global maximum value, we evaluate the limit as $x \to +\infty$:
$$ \lim_{x\to+\infty} g(x) = 1 - 1 - e^{-c/\sigma^2} + e^{-c/\sigma^2} = 0$$
So we have shown that $g(x) < 0$, (You may argue that if the equality hold, it will lead to contradiction, so the inequality here is strict) which is what we want.
Hopefully there are no silly calculation mistakes in the above steps.
A: Fairly easy proof for $c \geq 1$ and $\sigma^2=1$. There are famous inequalities for complementary cdf of standard normal distribution for positive $t$:
$$
\frac{1}{\sqrt{2\pi}}\frac{t}{t^2+1}e^{-t^2/2} \leq  \Phi^c(t) \leq \frac{1}{t}\frac{1}{\sqrt{2\pi}}e^{-t^2/2}.
$$
Proof can be found in Upper and lower bounds for the
normal distribution function by John D. Cook.
Therefore,
$$
\frac{\Phi^c(x+c/x)}{\Phi^c(x)}  \leq  \frac{\frac{1}{x+c/x}e^{-(x+c/x)^2/2}}{\frac{x}{x^2+1}e^{-x^2/2}} = \frac{e^{-c-(c/x)^2/2}}{ \frac{x^2+c}{x^2+1}}<e^{-c},
$$
which is true for $c \geq 1$.
A: \begin{align*} P(X>x+\epsilon) &= (2\pi \sigma^2)^{-1/2} \int_{x+\epsilon}^\infty e^{-y^2/2\sigma^2}dy\\
& = (2\pi \sigma^2)^{-1/2} \int_x^\infty e^{-(y+\epsilon)^2/2\sigma^2} dy\\
& = (2\pi \sigma^2)^{-1/2} \int_x^\infty e^{-\frac{ y^2 +2\epsilon y + \epsilon^2}{2\sigma^2}}dy \\
& \le e^{-\epsilon x /\sigma^2} e^{-\epsilon^2/2\sigma^2}P(X>x). \end{align*}
Setting $\epsilon = c/x$, we obtain 
$$P(X>x+c/x|X>x) \le e^{-c/\sigma^2} e^{-c^2/2x^2\sigma^2}.$$
