> For the complete documentation index, see [llms.txt](https://zavalab.gitbook.io/statschemebook/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://zavalab.gitbook.io/statschemebook/chapter-3-multivariate-random-variables/06-conditional-events.md).

# Conditional Events

* We have focused on probabilities conditional on simple events (e.g., $$X\_1=x\_1$$, $$X\_2\leq x\_2$$).
* Probabilities conditional on general events can be derived from domain properties.
* Return to definition of conditional probability:

  $$
  \begin{aligned}
  \mathbb{P}(X\in \mathcal{A}\_1|X\in \mathcal{A}\_2)=\frac{\mathbb{P}(X\in\mathcal{A}\_1;&; X\in \mathcal{A}\_2)}{\mathbb{P}(X\in \mathcal{A}\_2)}.
  \end{aligned}
  $$

  where $$\mathcal{A}\_1$$ and $$\mathcal{A}\_2$$ are subdomains of $$\mathcal{D}$$ that capture specific events.
* $$\mathbb{P}(X\in \mathcal{A}\_1|X\in \mathcal{A}\_2)$$ is prob that event $$\mathcal{A}\_1$$ occurs given that $$\mathcal{A}\_2$$ has occurred.
* $$\mathbb{P}(X\in\mathcal{A}\_1;&; X\in \mathcal{A}\_2)$$ is joint prob of $$\mathcal{A}\_1$$ and $$\mathcal{A}\_2$$, and $$\mathbb{P}(X\in \mathcal{A}\_2)$$ is marginal prob.
* For continuous $$X$$, prob $$\mathbb{P}(X\in \mathcal{A}\_1|X\in \mathcal{A}\_2)$$ is:

  $$
  \mathbb{P}(X\in \mathcal{A}\_1|X\in \mathcal{A}*2)=\frac{\displaystyle\int*{x\in \mathcal{A}\_1\cap\mathcal{A}*2}f(x)dx}{\displaystyle\int*{x\in \mathcal{A}\_2}f(x)dx}.
  $$
* Recall that joint event $${X\in\mathcal{A}\_1;&; X\in \mathcal{A}\_2}$$ is equivalent to $$X\in \mathcal{A}\_1\cap \mathcal{A}\_2$$.
* Conditional prob $$\mathbb{P}(X\_1=x\_1|X\_2=x\_2)$$ is a special case with events of the form $$\mathcal{A}\_1={X\_1=x\_1;&; X\_2\in \mathcal{D}\_2}$$ and $$\mathcal{A}\_2={X\_1\in \mathcal{D}\_1; & ; X\_2=x\_2}$$.
* Recall that, if $$\mathcal{A}\_1$$ and $$\mathcal{A}\_2$$ are *independent* events, then:

  $$
  \begin{aligned}
  \mathbb{P}(X\in\mathcal{A}\_1;&; X\in \mathcal{A}\_2)=\mathbb{P}(X\in \mathcal{A}\_1)\cdot \mathbb{P}(X\in \mathcal{A}\_2).
  \end{aligned}
  $$
* This implies that:

  $$
  \int\_{x\in \mathcal{A}\_1\cap\mathcal{A}*2}f(x)dx=\left(\int*{x\in \mathcal{A}*1}f(x)dx\right)\cdot \left(\int*{x\in \mathcal{A}\_2}f(x)dx\right)
  $$
* This explains why joint cdf is product of marginal cdfs for $$X\_1,X\_2$$ independent.
* Probability that event $$\mathcal{A}\_1$$ *or* $$\mathcal{A}\_2$$ occurs is given by:

  $$
  \mathbb{P}(X\in \mathcal{A}\_1\cup \mathcal{A}\_2)=\mathbb{P}(X\in \mathcal{A}\_1)+\mathbb{P}(X\in \mathcal{A}\_2)- \mathbb{P}(X\in \mathcal{A}\_1\cap \mathcal{A}\_2).
  $$
* This is computed by using joint and marginal pdfs of $$X$$ as:

  $$
  \mathbb{P}(X\in \mathcal{A}\_1\cup \mathcal{A}*2)=\int*{x\in \mathcal{A}*1}f(x)dx+\int*{x\in \mathcal{A}*2}f(x)dx- \int*{x\in \mathcal{A}\_1\cap\mathcal{A}\_2}f(x)dx.
  $$
* Pdfs and event properties can be used to compute probs of complex events.

## Example: Lifetime of Thermostat

{% hint style="info" %}
Compute probability that sensor or processor lasts more than a year.
{% endhint %}

* We build events $$\mathcal{A}\_1={x\_1\geq 1;&; x\_2\in \mathcal{D}\_2}$$ and $$\mathcal{A}\_2={x\_1\in \mathcal{D}\_1;& ;x\_2\geq 1}$$, which give $$\mathcal{A}\_1\cap\mathcal{A}\_2={x\_1\geq 1; & ; x\_2\geq 1}$$, and $$\mathcal{A}\_1\cup\mathcal{A}\_2={x\_1\geq 1; \textrm{or} ; x\_2\geq 1}$$.
* Computing probabilities we have:

  $$
  \begin{aligned}
  \int\_{x\in \mathcal{A}*1}f(x)dx&=\int*{0}^{\infty}\int\_{1}^{\infty}f(x\_1,x\_2)dx\_1dx\_2=0.37\\
  \int\_{x\in \mathcal{A}*2}f(x)dx&=\int*{1}^{\infty}\int\_{0}^{\infty}f(x\_1,x\_2)dx\_1dx\_2=0.90\\
  \int\_{x\in \mathcal{A}*1\cap\mathcal{A}*2}f(x)dx&=\int*{1}^{\infty}\int*{1}^{\infty}f(x\_1,x\_2)dx\_1dx\_2=0.33.
  \end{aligned}
  $$

  and thus $$\mathbb{P}(X\in \mathcal{A}\_1\cup \mathcal{A}\_2)=0.37+0.90-0.33=0.94$$.
* There is a high probability that either one of them lasts more than a year.

#### Python implementation: Union of events

$$\mathbb{P}(X\in\mathcal{A}\_1\cup\mathcal{A}\_2)=\mathbb{P}(X\in\mathcal{A}\_1)+\mathbb{P}(X\in\mathcal{A}\_2)-\mathbb{P}(X\in\mathcal{A}\_1\cap\mathcal{A}\_2)$$ with $$\mathcal{A}\_1={x\_1\geq 1}$$ and $$\mathcal{A}\_2={x\_2\geq 1}$$.

```python
P_A1 = P_box(1, np.inf, 0, np.inf)
P_A2 = P_box(0, np.inf, 1, np.inf)
P_A1_and_A2 = P_box(1, np.inf, 1, np.inf)
P_A1_or_A2 = P_A1 + P_A2 - P_A1_and_A2
print(f"P(A1) = {P_A1:.2f}, P(A2) = {P_A2:.2f}, P(A1 & A2) = {P_A1_and_A2:.2f}")
print(f"P(A1 or A2) = {P_A1_or_A2:.2f}")

# Monte Carlo check with S samples from the joint pdf (independent exponentials)
np.random.seed(0)
S = 100_000
x_omega = np.column_stack([np.random.exponential(mu_1, S), np.random.exponential(mu_2, S)])
P_hat = np.mean((x_omega[:, 0] >= 1) | (x_omega[:, 1] >= 1))
print(f"Empirical estimate P_hat(A1 or A2) = {P_hat:.2f}")
```

```
P(A1) = 0.37, P(A2) = 0.90, P(A1 & A2) = 0.33
P(A1 or A2) = 0.94
Empirical estimate P_hat(A1 or A2) = 0.94
```

**Note:** The joint pdf answers all questions about the thermostat: joint and marginal probabilities, logic of events, and expected lifetimes; Monte Carlo confirms the analytical results.