Question

Forensic analysis of JFK assassination bullets. Following theassassination of President John F. Kennedy (JFK) in 1963, the House Select Committee on Assassinations (HSCA) conducted an official government investigation. The HSCA concluded that although there was a probable conspiracy involving at least one shooter in addition to Lee Harvey Oswald, the additional shooter missed all limousine occupants. A recent analysis of assassination bullet fragments, reported in the Annals of Applied Statistics(Vol. 1, 2007), contradicted these findings, concluding that the evidence used by the HSCA to rule out a second assassin is fundamentally flawed. It is well documented that at least two different bullets were the source of bullet fragments found after the assassination. Let E= {bullet evidence used by the HSCA}, T= {two bullets used in the assassination}, and= {more than two bullets used in the assassination}. Given the evidence (E), which is more likely to have occurred— two bullets used (T) or more than two bullets used ?

a. The researchers demonstrated that the ratio,$\mathbf{P}\left(T\backslash E\right)\mathbf{/}\mathbf{P}\left(T^c\backslash E\right)$, is less than 1. Explain why this result supports the theory of more than two bullets used in the assassination of JFK.

b. To obtain the result, part a, the researchers first showed that $\frac{\mathbf{P}\left(T\backslash E\right)}{\mathbf{P}\left(T^c\backslash E\right)}\mathbf{=}\frac{\left[P\left(E\backslash T\right).P\left(T\right)\right]}{\left[P\left(E\backslash T^c\right).P\left(T^c\right)\right]}$Demonstrate this equality using Bayes’s Rule.

Short answer

a. As more than 2 bullets are likely to have been used in the execution.

b. The value is$\frac{\mathbf{P}\mathbf{(}\mathbf{T}\mathbf{\backslash }\mathbf{E}\mathbf{)}}{\mathbf{P}\mathbf{(}{\mathbf{T}}^{\mathbf{c}}\mathbf{\backslash }\mathbf{E}\mathbf{)}}\mathbf{=}\frac{[P\left(E\backslash T\right).P\left(T\right)]}{[P\left(E\backslash T^c\right).P\left(T^c\right)]}$.

Step-by-step solution

Important formula

The formula for probability is$\mathbf{P}\mathbf{=}\frac{\mathbf{favourableoutcomes}}{\mathbf{totaloutcomes}}$.

The Baye’s formula is

$$\begin{gathered} \mathbf{P}\mathbf{\left(}\frac{{\mathbf{B}}_{\mathbf{i}}}{\mathbf{A}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}\mathbf{⌒}\mathbf{A}\mathbf{)}}{\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{i}}\mathbf{\right)}\mathbf{P}\mathbf{\left(}{\displaystyle \frac{A}{B_i}}\mathbf{\right)}}{\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{1}}\mathbf{\right)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{i}}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{2}}\mathbf{\right)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{k}}\mathbf{\right)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{k}}}\mathbf{\right)}} \end{gathered}$$

Get the result that two bullets are likely to have been used in the execution

a.

$E\mathit{=}the\mathit{}HSCA\mathit{}relies\mathit{}on\mathit{}bullet\mathit{}evidence$

$T=theassas\sin ationwascarriedoutwithtwobullets$

$T^c=morethantwobulletsusedintheassassination.$

The require formula is $P\mathit{\left(}T\mathit{\right|}E\mathit{)}\mathit{<}P\mathit{(}{T}^{\mathit{c}}\mathit{\left|}E\mathit{\right)}$.

That is,

$\frac{P\left(T\right|E)}{P\left(T^c\right|E)}<1$

Thus, there is a greater chance that more than 2 bullets were used in the assassination than 2 bullets were used in the assassination based on the evidence.

Hence, the execution was likely to have involved more than two rounds.

Show the result of part (b).

b.

To establish the result

$P\left(T\right|E)/P(T^c\left|E\right)=\left[P\right(E\left|T\right).P\left(T\right)]/[P\left(E\right|T^c).P(T^c\left)\right]$

Now find the values are

$\begin{array}{l}P\left(T\right|E)=\frac{P\left(E|T\right)P\left(T\right))}{P\left(E\backslash T\right)P\left(T\right)}\\ P\left(T^c\backslash E\right)=\frac{P\left(E/T^c\right)P\left(T^c\right)}{P\left(E\backslash T\right)P\left(T\right)+\left(E\backslash T^c\right)P\left(T^c\right)}\end{array}$

Now,

$P\left(T\right|E)/P(T^c\left|E\right)=\frac{P\left(E\right|T)P\left(T\right)}{P\left(E\right|T^c)P\left(T^c\right)}$.

Therefore, the value is $P\left(T/E\right)/P\left(T^c\backslash E\right)=\frac{P\left(E/T\right)P\left(T\right)}{P\left(E\backslash T^c\right)P\left(T^c\right)}$.