Censored likelihoods

class: center, middle, inverse, title-slide

.title[
# Censored likelihoods
]
.author[
### Yue Jiang
]
.date[
### Duke University
]

---

### Customer service

Your last ten tickets took 2.5, 3.1, 5.7, 8.0, 9.8, 11.4, 15.5, 18.3, 24.1, and 26.9 minutes each to resolve.

.question[
- Suppose you assume resolution times are exponentially distributed. What is the probability a ticket takes less than 10 minutes? 
- What does the entire survival distribution look like?
- How does it compare to the Kaplan-Meier estimate?
]

---

### Estimation

Suppose we had a series of *non-censored* observations `\(t_1, t_2, \cdots, t_n\)`, 
and we thought that these survival times were i.i.d. and came from some
distribution with density `\(f(t)\)`.

.question[
How might we estimate the parameter(s) `\(\theta\)` of that distribution?
]

---

### Review: maximum likelihood estimation

`\begin{align*}
\mathcal{L}(\theta | T) &= f(t_1, t_2, \cdots, t_n | \theta)\\
&= f(t_1 | \theta)f(t_2 | \theta) \cdots f(t_n | \theta)\\
&= \prod_{i = 1}^n f(t_i|\theta),
\end{align*}`

which we can often maximize in closed form for many familiar distributions (or 
numerically).

---

### Review: maximum likelihood estimation

For instance, if we thought `\(T \stackrel{i.i.d}{\sim} Exp(\lambda)\)`, then

`\begin{align*}
\mathcal{L}(\lambda | T) &= \lambda^n \exp \left(-\lambda\sum_{i = 1}^n t_n \right)\\
\log \mathcal{L}(\lambda | T) &= n\log(\lambda) - \lambda \sum_{i = 1}^n t_i,\\
\hat{\lambda}_{MLE} &= \frac{n}{\sum_{i = 1}^n t_i}
\end{align*}`

---

### Back to customer service

Your last ten tickets took 2.5, 3.1, 5.7, 8.0, 9.8, 10+, 10+, 10+, 10+, and 10+ minutes each to resolve.

.question[
- What is the probability a ticket takes less than 10 minutes? 
- How would you estimate the survival function here (let's assume no parametric asusmption yet)
]

---

### Estimation for censored data

How might we perform maximum likelihood estimation for *censored* data? What
would the likelihood look like? Suppose we have `\(n\)` i.i.d. observations with the
same `\(f(t)\)`, `\(S(t)\)`, and hazard `\(\lambda(t)\)`. Consider what might happen at time
`\(t_i\)`.

.question[
Suppose an individual experiences event at `\(t_i: \delta_i = 1\)`. What do they contribute to the likelihood?

What would their contribution to the likelihood be at `\(t_i\)` if they were *censored*?
]

---

### Estimation for censored data

Remember that

`\begin{align*}
f(t) &= \lambda(t)S(t)
\end{align*}`

and so  an individual's contribution to the likelihood is thus

`\begin{align*}
&\mathrel{\phantom{=}} f(t_i)^{\delta_i}S(t_i)^{1 - \delta_i}\\
&= \lambda(t_i)^{\delta_i}S(t_i)^{\delta_i}S(t_i)^{1 - \delta_i}\\
&= \lambda(t_i)^{\delta_i}S(t_i)
\end{align*}`

---

### Estimation for censored data

For our exponential example with `\(T \stackrel{i.i.d.}{\sim} Exp(\lambda)\)`, since
the hazard is given by `\(\lambda\)` and the survival function is given by 
`\(e^{-\lambda t}\)`, we would thus have

`\begin{align*}
\mathcal{L}(\lambda | T) &= \prod_{i = 1}^n \lambda^{\delta_i}\exp(-\lambda t_i),
\end{align*}`

which we could then maximize using familiar methods.

---

### Back to customer service (again)

Your last ten tickets took 2.5, 3.1, 5.7, 8.0, 9.8, 10+, 10+, 10+, 10+, and 10+ minutes each to resolve.

.question[
- What is the probability a ticket takes less than 10 minutes assuming an exponential distribution for the resolution times?
]