# Show that $1-\Phi(x)$ is approximately $\varphi(x)/x$ for large $x$ (standard-normal random variable) [duplicate]

Demonstrate that, for a standard normal random variable:

$$1-\Phi(x) \approx \frac{\varphi(x)}x$$

for large values of $x$.

• – whuber Mar 8 '17 at 22:36

## 2 Answers

$1-\Phi(x)=\int_x^\infty \phi(t)\,dt$

$$=\int_x^\infty t^{-1}t\phi(t)\,dt$$

$u = t^{-1} , dv =t\phi(t)$ (since $\phi^{'}(t) =t\phi(t)$, $v=-\phi(t)$)

$$=-t^{-1}\phi(t)|_{t=x}^\infty - \int_x^\infty t^{-2}\phi(t)\,dt$$

$$=x^{-1}\phi(x) - \int_x^\infty t^{-2}\phi(t)\,dt$$

• So, $1-\Phi(x)=\frac{\phi(x)}{x} - \int_x^\infty t^{-2} \phi(t) \, dt < \frac{\phi(x)} x$ UPPER BOUND

Continuing with integration by parts,

$1-\Phi(x)=\phi(x)/x - \int_x^\infty t^{-2}\phi(t)\,dt$

$$=\phi(x)/x - \int_x^\infty t^{-3}t\phi(t)\,dt$$

$u=t^{-3}, dv = t\phi(t)$

$$=\phi(x)/x - \left[ -t^{-3}\phi(t)|_{t=x}^{\infty}- \int_x^\infty 3\cdot t^{-4} \phi(t) \, dt\right]$$

$$=\phi(x)/x - \left[ x^{-3}\phi(x) - 3\int_x^\infty t^{-4} \phi(t) \, dt\right]$$

$$=\phi(x)/x - x^{-3}\phi(x) + 3\int_x^\infty t^{-4}\phi(t) \, dt$$

• So, $1-\Phi(x)=\frac{\phi(x)}{x} - \frac{\phi(x)}{x^3} + 3\int_x^\infty t^{-4} \phi(t) \, dt > \frac{\phi(x)}{x} - \frac{\phi(x)}{x^3}$ LOWER BOUND

Thus, $$\frac{\phi(x)} x - \frac{\phi(x)}{x^3} < 1 - \Phi(x) < \frac{\phi(x)} x$$

$$\frac{\phi(x)} x \left( 1 - \frac{\phi(x)}{x^2} \right) < 1 - \Phi(x) < \frac{\phi(x)} x$$

$$\frac{\phi(x)} x ( 1 - o(x)) < 1 - \Phi(x) < \frac{\phi(x)} x$$

Therefore, by sandwich thrm, $1 - \Phi(x) \approx \frac{\phi(x)}{x}$ for large $x$.

Note that

$$x \left(1-\Phi(x) \right) \to 0, \quad \text{as}\quad x \to \infty$$

because the expectation exists. Applying L'Hopital's rule twice, we then get

\begin{align} \lim_{x\to\infty} \frac{x \left(1-\Phi(x) \right)}{\phi(x)} &= \lim_{x\to\infty} \frac{ 1-\Phi(x) - x \phi(x)}{-x\phi(x)} \\ &= 1 - \lim_{x\to \infty}\frac{1-\Phi(x)}{x \phi(x)} \\ & = 1 - \lim_{x \to \infty} \frac{-\phi(x)}{ \phi(x) -x^2 \phi(x)} \\ & = 1 + \lim_{x\to \infty} \frac{1}{1-x^2} \\ & =1 \end{align}

where we have used that $$\phi^{\prime} (x) = -x \phi(x)$$. Hence

$$1-\Phi(x) \sim \frac{\phi(x)}{x}$$

(their ratio goes to 1 for large values of $$x$$)