Strategic conditioning

Idea: Split by the fact that helps

Ask which single fact, if you were told it, would make the problem easy, then condition on it.

• A hidden state, such as a regime
• A random count, such as how many trades arrive
• The result of the first step

Formula: Total probability

$P(A)=\sum P(A\mid B=b)P(B=b)$

Add the easy cases, each with its weight; the cases cover everything once and the weights add to $1$.

Example: Breach by regime

Calm, stressed and crisis days have chances $50\%$, $30\%$, $20\%$; breach chances $1\%$, $5\%$, $20\%$. Chance of a breach?

1. Calm: $0.5\times 0.01=0.005$
2. Stressed: $0.3\times 0.05=0.015$; crisis: $0.2\times 0.2=0.04$
3. Total: $0.005+0.015+0.04=0.06$

Common mistake: Averaging the conditional chances

Writing $\frac{0.01+0.05+0.2}{3}\approx 0.087$ treats the regimes as equally likely and overstates the risk by nearly half.

Instead Weight by the base rates. If you cannot state $P(B=b)$, you chose the wrong variable.

Idea: Name the case that carries it

Crisis has the smallest weight yet carries $\frac{0.04}{0.06}=\frac{2}{3}$ of the risk.

• That is where a hedge should go
• Averages split the same way: $E[X]=\sum E[X\mid B]P(B)$
• A changed number moves only one line of work

Example: A random number of coins

A fair die with faces $1$ to $4$ shows $N$, then you toss $N$ fair coins. Chance of exactly $1$ head?

1. Given $N=m$, the chance is $\frac{\left(\genfrac{}{}{0ex}{}{m}{1}\right)}{2^m}$
2. For $m$ from $1$ to $4$: $\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}=\frac{13}{8}$
3. Each $N$ has weight $\frac{1}{4}$: $\frac{13}{8}\times \frac{1}{4}=\frac{13}{32}$

Formula: First-step analysis

$h_i=p{h}_{i+1}+(1-p)h_{i-1}$

After one step the same problem reappears from the new state; solve with the known ends.

Quick check: A walk with drift

From $2$, a walk steps up with chance $\frac{2}{3}$, down with $\frac{1}{3}$, and stops at $0$ or $4$. Chance it hits $4$ first?

Idea: Expected times work the same

Tosses until the first head: one toss always happens, then half the time you are back at the start.

• $E[T]=1+\frac{1}{2}\times E[T]$, so $E[T]=2$
• No infinite sum in sight
• A fair walk from the middle gives $\frac{2}{4}=\frac{1}{2}$; drift adds $\frac{4}{5}-\frac{1}{2}=\frac{3}{10}$

Idea: Say the split aloud

Name it before you compute: by regime, by count or by first step.

• Check the cases cover everything, weights add to $1$
• Work one case at a time
• Finish by naming the case that carries the answer