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Joint vs conditional probability

ABSMOGHE

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Hi I was trying to understand the difference between Joint Probability and Conditional Probability

I came across this post of yours.
https://www.bionicturtle.com/forum/threads/p2-t6-307-hazard-rate-malz-section-7-2.6932/#post-42375

conditional default probability in the 2nd year; i.e., the probability of default in year 2 conditional on survival through the end of the previous year
AND

Unconditional / Joint Default Probability Hull does refer to this as unconditional because it is "unconditional" as seen from Time zero,

What I do not understand is the difference between both. The wordings of both appear to be the same.
In this case what would Joint Probability exactly signify, and what would Conditional Probability signify ?
Also, what would this mean : "Probability of survival in first year followed by default in second year". Would this be joint probability or conditional probability ? If it means one of the above, how would you phrase the other one (ie. if the above term means joint probability, how would you phrase conditional probability as a question) ?

I apologise if the question was a foolish one. However I am having really hard time trying to grasp these concepts.

Thanks a lot.
 

David Harper CFA FRM

David Harper CFA FRM (test)
Staff member
#2
Hi @ABSMOGHE As this question (or variation on the same) has been asked so often, I've elaborated here in the text associated with our new YouTube video feature posts https://www.bionicturtle.com/forum/...l-vs-conditional-probability.21654/post-71326 i.e.,
The distinction between joint versus conditional versus unconditional probability is one of the most commonly asked questions on this forum! The video above illustrates these terms in the context of a probability matrix. To supplement this video I have prepared a very succinct spreadsheet here (https://www.dropbox.com/s/esvv53sm9vha1bv/102918-pd-conditional-joint-unconditional-etc.xlsx?dl=0); the screen shot is below.

You will notice that the sheet only contains a single input, given by λ = 3.0%, which is the conditional default probability (conditional PD); in continuous time, this is called a hazard rate (the hazard rate is an instantaneous conditional PD). So, the context here is default probabilities, obviously! But we can illustrate the three most common PD terms. I will use the discrete version:
  • if the annual PD = 3.0%, then the 5-year cumulative PD = 1 - (1 - 0.05)^5 = 14.13%. This means that there exists a 14.13% probability of a default during any of the years (i.e., cumulatively) before the end of the fifth year. We can also infer that the 5-year cumulative survival probability = 1 - 14.13% = 85.87%
  • The unconditional probability of default during the 5th year is given by: cumulative_PD (5 years) - cumulative_PD (4 years) = 14.13% - 11.47% = 2.66%. By unconditional, we mean that this is the probability of default during the 5th year as seen from today (which is time zero). This is in contrast to a default probability conditional on survival through the Xth (eg., 4th year); that is, it is not conditional on any survival, but rather from the initial perspective. Hopefully you can see how this is consistent with an unconditional Prob(A), as opposed to a conditional Pr(A | B).
  • Skipping to the bottom row, the conditional probability of default during the 5th year = unconditional_PD (5th year) / Cumulative_Survival_PD (through the end of the 4th year) = 2.66%/(1 - 11.47%) = 3.00%. It's no accident this equals our conditional PD input assumption! This probability is "conditional on" survival to the end of the prior year.
  • Now back to the third row: we can ask the question, what is the joint probability of survival to the end of the 4th year and (then) default during the 5th year? This is a straightforward application of the text: Joint PD (survive to end of 4th default during 5th year) = (1 - 11.47%)*3.0% = 2.66%. We just multiply the probability of survival by the probability of subsequent default. This gives us the same answer as the unconditional PD (i.e., 3.0%)! That is not a coincidence: in this context, such a joint probability is "from the original perspective of time zero," so it is also unconditional (as in, it does not condition on any prior survival assumptions). They are the same in this context, but we do want to distinguish conceptually between the unconditional Probability(A) and the Joint Probability (A ∩ B) because, in the probability matrix illustrated in the video above, they are not the same! The context matters ....


Related Q&A (Question and Answer) in this forum:
So to apply that explanation to your question, let us assume lambda λ = 3.0% is constant (as shown), for convenience sake. Further, I'll use discrete (annual) time, although it's essentially similar in continuous time.
  • The Joint Probability is the probability of survival (let's say to the end of the 2nd year) and default during the 3rd year. "Joint" means that both things happen in sequence, one then the other. This Joint Prob (survive to end of 2nd year and then default during 3rd year) = Prob (cumulative survival thru 2nd year) * conditional PD (during 3rd year | survival thru 2nd year) = 0.97^2 * 3.0% = 2.82%. But, in this context, this is also the unconditional probability of default during the 3rd year because it is "from the perspective of today, time zero" and therefore does not itself depend (does not condition) on an assumption about events prior to this third year. It is a probability that is "unconditional" in the sense that it is from today's (time zero) perspective.
  • But the Conditional PD during the 3rd year is 3.0%. Yes, it's an input, but it is also given by: Conditional_PD (during 3rd year) = Joint_PD (Survival thru 2nd year ∩ default during 3rd year) / Prob_Survival (thru the end of the 2nd year) = 2.82% / 97.0%^2 = 3.0%. This is the most basic relationship: Conditional Prob (A | B) * Unconditional Prob (B) = Joint Prob (A ∩ B), This 3.0% is not unconditional (i.e., not from the initial perspective) but rather is conditional on survival thru the end of the 2nd year; so we can view this as a "from the perspective of the beginning of the third year" (i.e., conditional on survival up to that point in time) rather than "from the initial--time zero--perspective" (i.e. unconditional on any subsequent survival assumptions). I hope that clarifies!
 
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#4
I understand mathematically why the conditional PD will be higher than the unconditional PD. But I have a hard time understanding WHY this is the case intuitively. For example, in mortality tables, your probability of dying in the next year if you reach 40 years old (i.e. conditional probability) is surely LOWER than your probability of dying in your 41st year taken at birth (unconditional) ? Why wouldn't this be the case for PD?
 

David Harper CFA FRM

David Harper CFA FRM (test)
Staff member
#5
@claudia99 but it's not, right? It acts the same in the mortality table ... the conditional probability of death during the 41st year must be higher than the corresponding unconditional probability. Say the cumulative prob of survival through the end of the 40th year is 96.0% and the conditional prob of default during the 41st year is 0.20%. Then the unconditional (aka, joint) prob of death during the 41st years = 96.0% * 0.20% = 0.1920%; the unconditional must be lower mathematically due to the cumulative survival probability, as multiplier, must be less than 1.0.

Intuitively, at least the way I look at is, an unconditional probability of death (e.g., during the 41th year) is a narrow outcome (or event) within a much wider probability space (including all 40 prior years before). Thanks!
 
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