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Crosspost from math.stacksexchange. Though it might fit better here.

My question is about the possibility of showing equivalence between the hazard rate, the conditional probability (of failure) and a likelihood function.


Dynamics: Consider a coin that has $P(\boldsymbol{H})=p$ and $P(\boldsymbol{T})=1-p$ such that $P(H) + P(T) = 1$. The stochastic process $X(t)$ evolves according to this dynamic over time $t$ resulting in a series of $\boldsymbol{H}$ until at time $T$ it lands on tails $\boldsymbol{T}$ which ends the sequence.


Definitions:

  • Consider $T$ a continuous random variable with density $f(t)$ and cumulative density $F(t)$.

1) Hazard rate: $\mu(t)=\frac{f(t)}{1-F(t)}$.

Aka. force of mortality, hazard function, failure rate, see wiki. This can be larger than one.

2) Conditional probability of failure: $P(t<T\leq t+\triangle t|T\geq t)=\frac{P(t<T\leq t+\triangle t)\cap P(T>t)}{P(T>t)}\approx\mu(t)\triangle t$

Showing equivalence between 1) and 2) for $\triangle t\rightarrow0$ is easy. Again, see wiki. 2) is the conditional probability of failure, given that the unit worked up until time $t$.

3) Likelihood function: $\mathcal{L}(\theta|X)$.

Where $\theta$ is a parameter ($p$) and $X$ is the data observed (e.g. the series of H). E.g. assuming a fair coin we can write $\mathcal{L}(P(\boldsymbol{H})=0.5|\boldsymbol{HH})=P(\boldsymbol{HH}|P(\boldsymbol{H})=0.5)=0.25$. Or alternatively, $P(\boldsymbol{T}|\boldsymbol{HH})=1-P(\boldsymbol{HH}|P(\boldsymbol{H})=0.5)=0.75.$


Questions:

q. 1) Am I right in asserting that the likelihood function $\mathcal{L}(\theta|X)$ $is$ the complement to the conditional probability function $P(t<T\leq t+\triangle t|T\geq t)$?

Exemplified as $P(\boldsymbol{T}|\boldsymbol{HH})=1-P(\boldsymbol{HH}|P(\boldsymbol{H})=0.5)=0.75.$

q. 1.2) If q .1 is true, how is the likelihood function connected to the hazard rate function? Can they be shown to be equivalent?

q. 1.3) Intuitively, all conditional probabilities are purely multiplicative processes. Can we generally write this as $P(t<T\leq t+\triangle t|T\geq t) = 1-\prod^{t+\triangle t}_tP(\boldsymbol{H})_t$? E.g. as the conditional probability of realising another $\boldsymbol{H}$ decreases, the probability (and thus likelihood?) of realising $\boldsymbol{T}$ increases.

q. 2) Is it generally true, that if all conditional probablities are purely multiplicative processes of numbers between one and zero, they all decrease with some exponential rate $\lambda$?

Consider the cumulative hazard rate: $\varGamma=\int_0^t\mu(u)du=-log(1-F(t))$. See wiki.

q. 2.1) If q. 2 is correct, and we can express the integral over hazard rate as the negative logarithm to the complement of the cdf $-log(1-F(t)$ (sometimes called the survival function), is this essentially the same as the conditional probability?

q. 3) How does the above relate to the notion of sojourn time? One can imagine $N$ states with $N$ different hazard rates. What is this problem generally called and how can I relate it to expected mean sojourn time? Intuitively, states with a high hazard rate will have lower sojorn time. Are these equivalent somehow?

q. 4) If anyone would care to relate this to a Markov Chain with absorbing states, they are most welcome. Especially I am interested in how the conditinal probability relates to the transition probabilities.


I realise these are a lot of questions. Any hints, partial answers or call for clarity is appreciated, thanks!

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My question is about the possibility of showing equivalence between the hazard rate, the conditional probability (of failure) and a likelihood function.

TLDR; There is no such equivalence.

Likelihood is defined as

$$ \mathcal{L}(\theta \mid x_1,\dots,x_n) = \prod_{i=1}^n f_\theta(x_i) $$

so it is a product of probability density functions evaluated at $x_i$ points, given some fixed value of parameter $\theta$.

So it has nothing to do with hazard rate, since hazard rate is probability density function evaluated at $x_i$ point parametrized by $\theta$, divided by survival function evaluated at $x_i$ parametrized by $\theta$

$$ h(x_i) = \frac{f_\theta(x_i)}{1-F_\theta(x_i)} $$

Moreover, likelihood is not a probability and it is not a conditional probability. It is a conditional probability only in Bayesian understanding of likelihood, i.e. if you assume that $\theta$ is a random variable.

Your understanding of conditional probability also seems to be wrong:

Intuitively, all conditional probabilities are purely multiplicative processes. [...] Is it generally true, that if all conditional probablities are purely multiplicative processes of numbers between one and zero, they all decrease with some exponential rate $\lambda$?

The answer is no. Conditional probability is a relation between joint and individual probabilities

$$ P(A \mid B) = \frac{P(A \cap B)}{P(B)} $$

So it is not a process, and it is not multiplicative. Moreover, multiplicative relation is a definition of independence

$$ P(A \cap B) = P(A)\,P(B) $$

or equivalently

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

Even if you are talking about a random process, then this is not true. To give an example, imagine a series of coin tosses, if they are independent, then probability of tossing head given that previous toss resulted in tails is simply a (unconditional) probability of tossing head.

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  • $\begingroup$ Thanks for the answers! 1) If, in the Bayesian sense (e.g. $\theta$ is a rv), the likelihood function is a conditional probability, does this link to the conditional probability of failure as defined in 2)? 2) Am I correct in asserting that (in a coin toss experiment) if $Pr(H)=0.5$ then the conditional probability (conditioned on two heads) is $Pr(H|HH)=0.5*0.5*0.5$? How is this not a multiplicative process? 3) Is the folliwng statement true? Any process that multiplies numbers together between 0 and 1 will decay with some rate over time as $e^{-\lambda t}$? $\endgroup$ – tmo Dec 2 '16 at 13:10
  • $\begingroup$ 1) Obviously $\prod_i f(x_i) \ne \frac{f(x_i)}{1 - F(x_i)}$; 2) for i.i.d. coin tosses $P(A|B)=P(A)$, so the conditional probability is 0.5; and 3) is a different question, deserting it's own question rather a comment. $\endgroup$ – Tim Dec 2 '16 at 13:17

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