Probability and Non-Probability Sampling
Investors make investment decisions based on their experience and expertise. Their decisions may change in the wake of new knowledge and observations.
Bayes’ formula allows us to update our decisions as we receive new information. In other words, it calculates an updated or posterior probability given a set of prior probabilities for a given event.
Given a set of prior probabilities for an event, if we receive new information, the updated probability is as follows:
$$ \begin{align*}\text{Updated probability of an event given the new information}
= & \frac {\text{Probability of the new information given event}}{\text{Unconditional probability of the new information}} \\ & \times {\text{Prior probability of event.}}
\end{align*} $$
The above equation can be written as:
$$ P(\text{Event}\mid \text{Information})=\frac { P(\text{Information}\mid \text{Event})}{P(\text{Information})}\cdot P(\text{Event}) $$
Let \(B_1, B_2, B_3,\ldots, B_n\) be a set of mutually exclusive and exhaustive events.
Using the conditional probability:
$$ P(B_i \mid A)=\frac {P(B_i\cap A)}{P(A)} \ldots \ldots (1) $$
And also, the relationship:
$$ P(B_i \cap A)=P(A \cap B_i )=P(B_i )\cdot P(A\mid B_i ) \ldots \ldots (2)$$
Also, using the total probability rule:
$$ P(A)=\sum_{i=1}^n P(A \cap B_i) =\sum_{i=1}^n P(B_i) \cdot P(A\mid B_i ) \ldots \ldots (3) $$
Substituting equations (2) and (3) in (1), we have:
$$ P(B_i \mid A)=\frac {(P(A \mid B_i )}{\sum_{i=1}^n P(B_i)\cdot P(A\mid B_i)} \cdot P(B_i) $$
This is the Bayes’ formula, and it allows us to ‘turnaround’ conditional probabilities, i.e., we can calculate \(P(B_i \mid A)\) if given information only about \(P(A\mid B_i)\).
Note that:
Example: Bayes’ Formula
An Investment Analyst wishes to investigate a stock’s performance by considering a number of stocks listed on different exchanges. In the sample, 50% of stocks were listed on the New York Stock Exchange (NYSE), 30% on the London Stock Exchange (LSE), and 20% on the Tokyo Stock Exchange (TSE).
The probability of a stock posting a negative return on the NYSE, LSE, and TSE is 40%, 35%, and 25%, respectively.
If the Analyst picks a stock at random from this group, what is the probability that it has a negative return on the NYSE?
Solution
We are looking for P(NYSE | Negative Return).
Let’s define the following events:
NYSE is the event “A stock chosen at random is listed on the NYSE.”
LSE is the event “A stock chosen at random is listed on the LSE.”
TSE is the event “A stock chosen at random is listed on the TSE.”
Finally, let NR be the event, “A randomly chosen stock posts a negative return.”
Therefore,
$$ \begin{align*}
P\left(NYSE\middle|NR\right)& =\frac{P\left(NYSE\right)P(NR|NYSE)}{P\left(NYSE\right)P\left(NR\middle|NYSE\right)+P\left(LSE\right)P\left(NR\middle|LSE\right)+P\left(TSE\right)P(NR|TSE)}\\
& =\frac {0.5\times 0.4}{0.5\times 0.4+0.3 \times 0.35+0.2 \times 0.25} \\
&=\frac {0.2}{0.355} \\
& =0.5634\approx 56.3\%
\end{align*} $$
Question
You have developed a set of criteria for assessing potential investments in growth-stage companies. Companies not meeting these criteria are predicted to be insolvent within 24 months. You gathered the following information when validating your criteria:
- Fifty percent of the companies that have been assessed will become insolvent within 24 months: \(P(\text{insolvency}) = 0.50\).
- Sixty-five percent of the companies assessed meet the criteria: \(P(\text{meet criteria}) = 0.65\).
- The probability that a company will meet the criteria given that it remains solvent for 24 months is 0.80: \(P(\text{meet criteria} \mid \text{solvency}) = 0.80\).
The probability that a company will remain solvent, given that it meets the criteria, that is, \(P(\text{solvency} \mid \text{meet criteria})\), is closest to:
- 20%.
- 50%.
- 62%.
Solution
Using Bayes’ formula, we have:
$$ \begin{align*}
& P(\text{solvency} \mid \text{meet criteria})\\ & =\frac {P(\text{meet criteria}\mid \text{solvency})P(\text{solvency})}{{ [P(\text{meet criteria}\mid \text{solvency})P(\text{solvency})} \\ {+P(\text{meet criteria}\mid \text{insolvency})P(\text{insolvency}) ] }} \\
&=\frac {0.80 \times 0.50}{0.80 \times 0.50+P(\text{meet criteria}\mid \text{insolvency})\times 0.50} \\
\end{align*} $$Clearly, we need to calculate the \(P(\text{meet criteria}\mid \text{insolvency})\). Using the total probability:
$$ \begin{align*}
P(\text{meet criteria}) & = P(\text{meet criteria} \mid \text{solvency})P(\text{solvency}) \\ & + P(\text{meet criteria} \mid \text{insolvency}) P(\text{insolvency}) \\
\Rightarrow 0.65 & = 0.80 \times 0.50 + P(\text{meet criteria} \mid \text{insolvency})\times 0.50
\end{align*} $$$$ \therefore P(\text{meet criteria} \mid \text{insolvency})=\frac {0.65-0.80 \times 0.50}{0.50}=0.50 $$
As such,
$$ P(\text{solvency} \mid \text{meet criteria})=\frac {0.80 \times 0.50}{0.80 \times 0.50+0.50 \times 0.50}=\bf{0.6153}\approx 62\% $$