Calculus: taking a derivative of a cdf

Question

Let $W_t$ be a standard Wiener process and $\tau_x (x > 0)$ be the first passage time to level $x (\tau_x = min \{t; W(t) = x\})$ .

What are the probability density function and the expected value of $\tau_x$?

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My attempt

$P(\tau_x \leq t) = P(\tau_x \leq t, w_t < x) + P(\tau_x \leq t, w_t > x) = 2 \int_{x}^{\infty} \frac{1}{\sqrt{2\pi t}} e ^{-\frac{x^2}{2t}} dx$

set $v = \frac{x}{\sqrt{t}}$
then $dv = \frac{1}{\sqrt{t}}dx$

$P(\tau_x \leq t) = 2 \int_{v}^{\infty} \frac{1}{\sqrt{2\pi}} e ^{-\frac{v^2}{2}} dv $

Then I am stuck at the following. I am not quite sure how to take the probability with respect to t

$\frac{dP(\tau_x \leq t)}{dt} = $

Thank you for your time.

Answer 1

Let $x\geq 0$ and $t\geq 0$. Note that by symmetry $P(W_t \geq x | \tau_x \leq t) = 1/2$. Also $P(W_t\geq x) = P(W_t\geq x |\tau_x\leq t)P(\tau_x\leq t) + P(W_t\geq x |\tau_x> t)P(\tau_x> t) = P(W_t\geq x |\tau_a\leq t)P(\tau_x\leq t)+0= P(W_t\geq x |\tau_a\leq t)P(\tau_x\leq t)$.

Hence $P(\tau_x \leq t) = 2P(W_t\geq x)$, you can calculate the last probability and so the distribution and density of $\tau_x$.

Answer 2

By the reflection principle, the area under and above $x$ is equivalent. Therefore, by the law of total probability we have

$P(\tau_x \leq t) = P(\tau_x \leq t, w_t < x) + P(\tau_x \leq t, w_t > x) = 2P(\tau_x \leq t, w_t > x) = 2P(w_t > x) = 2\int_{x}^{\infty} \frac{1}{\sqrt{2\pi t}} e ^{-\frac{w^2}{2t}} dw$

Note that my initial attempt is incorrect to take the integral with respect to x. The integral needs to be taken with respect to $w$ here, which is defined as $W_t = w$. When $W_t = w$, note $t$ is a constant.

set $v = \frac{w}{\sqrt{t}}$, $dv = \frac{1}{\sqrt{t}}dw$

Substitute $v$ into the equation,

$P(\tau_x \leq t) = 2 \int_{v}^{\infty} \frac{1}{\sqrt{2\pi}} e ^{-\frac{v^2}{2}} dv $

To change the limit into $t$, substitute $x$ into the function v so that the results become $ v = \frac{w}{\sqrt{t}} = \frac{x}{\sqrt{t}}$

$P(\tau_x \leq t) = 2 \int_{\frac{x}{\sqrt{t}}}^{\infty} \frac{1}{\sqrt{2\pi}} e ^{-\frac{v^2}{2}} dv = 2(1-\int_{-\infty}^{\frac{x}{\sqrt{t}}} \frac{1}{\sqrt{2\pi}} e ^{-\frac{v^2}{2}} dv) = 2 - 2N(\frac{x}{\sqrt{t}})$

where $N(\frac{x}{\sqrt{t}})$ denotes the cdf of a normal distribution up to $\frac{x}{\sqrt{t}}$.

Now, we take the derivative of $P(\tau_x \leq t)$ with respect to $t$. This can be broken down into two parts, $\frac{dP(\tau_x \leq t)}{d(\frac{x}{\sqrt{t}})} $ and $\frac{d\frac{x}{\sqrt{t}}}{dt}$

$\frac{dP(\tau_x \leq t)}{dt} = \frac{dP(\tau_x \leq t)}{d(\frac{x}{\sqrt{t}})}\frac{d\frac{x}{\sqrt{t}}}{dt} = \frac{d(2 - 2N(\frac{x}{\sqrt{t}}))}{d\frac{x}{\sqrt{t}}}\frac{d\frac{x}{\sqrt{t}}}{dt}=-2N'(\frac{x}{\sqrt{t}}) \times (-\frac{1}{2}xt^{-\frac{3}{2}}) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(x/t)^2}{2}}$

The expected value of $\tau_x$ is essentially the drunk man problem. The expectation is composed of the probability of reaching $x$ or $-\infty$, so

$P_x \cdot x + (1-P_x) \cdot \infty = 0 $

which yields

$P_x = \infty $