const hazard rate, with graph

label – intuitive, mathStat

 

Q1: Intuitively, how is const hazard rate different from constant density i.e. uniform distro?

 

It's good to first get a clear (hopefully intuitive) grasp of constant hazard rate before we talk about the general hazard rate. I feel a common usage of hazard rate is in the distribution of lifespan i.e. time-to-failure (TTF).

 

Eg: run 999999 experiments (what experiment? LG unimportant) and plot histogram of the lifespan of ..... Intuitively, you won't see a bunch of bars of equal height – no uniform distro!

Eg: 10% of the remaining (poisonous!) mercury evaporates each year, so we can plot histogram of lifespan of mercury molecules...

Eg: Hurricane hit on houses (or bond issuers). 10% of the remaining shanties get destroyed each year...

Eg: 10% of the remaining bonds in a pool of bonds default each year. Histogram of lifespan ~= pdf graph...

 

 

If 10% of the survivors fail each year exactly, there's not much randomness here:) but let's say we have only one shanty named S3, and each year there's a 10% chance of hazard (like Hurricane). The TTF would be a random variable, complete with its own pdf, which (for constant hazard rate) is the exponential distribution. As to the continuous case, imagine that each second there's a 0.0000003% chance of hazard i.e. 10% per year spread out to the seconds...

 

I feel there are 2 views in terms of noisgen. You can say the same noisegen runs once a year, or you can say for that one shanty (or bond) we own, at time of observation, the noisegen runs once only and generates a single output representing S3's TTF, 0 < TTF < +inf.

 

How does the e^{- λt} term come about? Take mercury for example, starting with 1 kilogram of mercury, how much is left after t years? Taking t = 3, it's (1-10%)^3. In other words, cumulative probability of failure = 1- (1-10%)^3. Now divide each year into n intervals. Pr(TTF < t) = 1- (1- 10%/n) ^ n*t. As n goes to infinity, Pr(TTF < t years) = 1- e^{- 0.1t}  i.e. the exponential distribution.

 

(1 - 0.1/n)ⁿ approaches e^{- 0.1}     as n goes to infinity.

This is strikingly similar to 10%/year continuous compounding

 

(1 + 0.1/n)ⁿ approaches e^{+ 0.1}     as n goes to infinity.

 

A1: Take the shanty case. Each year, the same number of shanties collapse -- uniform density, but as the survivor population shrinks, the chance of failure becomes very high.