Probabilidad condicional

Si${\displaystyle P(B)=0}$, entonces según la definición,${\displaystyle P(A\mid B)}$no está definido .

El caso de mayor interés es el de una variable aleatoria Y , condicionada a una variable aleatoria continua X que resulta en un resultado particular x . El evento${\displaystyle B=\{X=x\}}$tiene probabilidad cero y, como tal, no se puede condicionar.

En lugar de condicionar a que X sea exactamente x , podríamos condicionar a que esté más cerca que la distancia${\displaystyle \varepsilon }$lejos de x . El evento${\displaystyle B=\{x-\varepsilon <X<x+\varepsilon \}}$En general, tendrá una probabilidad distinta de cero y, por lo tanto, se puede condicionar. Entonces podemos tomar el límite.

Por ejemplo, si dos variables aleatorias continuas X e Y tienen una densidad conjunta${\displaystyle f_{X,Y}(x,y)}$, luego por la regla de L'Hôpital y la regla integral de Leibniz , al diferenciar con respecto a${\displaystyle \varepsilon }$:

${\displaystyle {\begin{aligned}\lim _{\varepsilon \to 0}P(Y\in U\mid x_{0}-\varepsilon <X<x_{0}+\varepsilon )&=\lim _{\varepsilon \to 0}{\frac {\int _{x_{0}-\varepsilon }^{x_{0}+\varepsilon }\int _{U}f_{X,Y}(x,y)\,\mathrm {d} y\,\mathrm {d} x}{\int _{x_{0}-\varepsilon }^{x_{0}+\varepsilon }\int _{\mathbb {R} }f_{X,Y}(x,y)\,\mathrm {d} y\,\mathrm {d} x}}\\[6pt]&={\frac {\int _{U}f_{X,Y}(x_{0},y)\,\mathrm {d} y}{\int _{\mathbb {R} }f_{X,Y}(x_{0},y)\,\mathrm {d} y}}.\end{aligned}}}$

El límite resultante es la distribución de probabilidad condicional de Y dado X y existe cuando el denominador, la densidad de probabilidad${\displaystyle f_{X}(x_{0})}$, es estrictamente positivo.

Resulta tentador definir la probabilidad indefinida.${\displaystyle P(A\mid X=x)}$utilizando limit( 1 ), pero esto no se puede hacer de manera consistente. En particular, es posible encontrar variables aleatorias X y W y valores x , w tales que los eventos${\displaystyle \{X=x\}}$y${\displaystyle \{W=w\}}$son idénticos, pero los límites resultantes no lo son:

${\displaystyle \lim _{\varepsilon \to 0}P(A\mid x-\varepsilon \leq X\leq x+\varepsilon )\neq \lim _{\varepsilon \to 0}P(A\mid w-\varepsilon \leq W\leq w+\varepsilon ).}$

La paradoja de Borel-Kolmogorov lo demuestra mediante un argumento geométrico.

Sea X una variable aleatoria discreta y sus posibles resultados denotemos V. Por ejemplo, si X representa el valor de un dado lanzado, entonces V es el conjunto${\displaystyle \{1,2,3,4,5,6\}}$Supongamos, a efectos de presentación, que X es una variable aleatoria discreta, de modo que cada valor en V tiene una probabilidad distinta de cero.

Para un valor x en V y un evento A , la probabilidad condicional viene dada por${\displaystyle P(A\mid X=x)}$. Escribiendo

${\displaystyle c(x,A)=P(A\mid X=x)}$

En resumen, vemos que es una función de dos variables, x y A.

Para un A fijo , podemos formar la variable aleatoria${\displaystyle Y=c(X,A)}$Representa un resultado de${\displaystyle P(A\mid X=x)}$siempre que se observe un valor x de X.

La probabilidad condicional de A dado X puede tratarse, por lo tanto, como una variable aleatoria Y con resultados en el intervalo${\displaystyle [0,1]}$. Según la ley de probabilidad total , su valor esperado es igual a la probabilidad incondicional de A.

La probabilidad condicional parcial ${\displaystyle P(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m})}$ se trata de la probabilidad de un evento ${\displaystyle A}$dado que cada uno de los eventos de condición ${\displaystyle B_{i}}$ha ocurrido en cierta medida ${\displaystyle b_{i}}$(grado de creencia, grado de experiencia) que podría ser diferente del 100%. Frecuentísticamente, la probabilidad condicional parcial tiene sentido si las condiciones se prueban en repeticiones experimentales de duración apropiada. ${\displaystyle n}$. ^{[ 10 ]} Tal ${\displaystyle n}$La probabilidad condicional parcial acotada se puede definir como la ocurrencia promedio esperada condicionalmente del evento.${\displaystyle A}$en bancos de prueba de longitud ${\displaystyle n}$que se ajustan a todas las especificaciones de probabilidad ${\displaystyle B_{i}\equiv b_{i}}$, es decir:

${\displaystyle P^{n}(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m})=\operatorname {E} ({\overline {A}}^{n}\mid {\overline {B}}_{1}^{n}=b_{1},\ldots ,{\overline {B}}_{m}^{n}=b_{m})}$^{[ 10 ]}

Sobre esa base, la probabilidad condicional parcial se puede definir como

${\displaystyle P(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m})=\lim _{n\to \infty }P^{n}(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m}),}$

dónde${\displaystyle b_{i}n\in \mathbb {N} }$^{[ 10 ]}

La condicionalización de Jeffrey ^{[ 11 ] [ 12 ]} es un caso especial de probabilidad condicional parcial, en el que los eventos de condición deben formar una partición :

${\displaystyle P(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m})=\sum _{i=1}^{m}b_{i}P(A\mid B_{i})}$

When Morse code is transmitted, there is a certain probability that the "dot" or "dash" that was received is erroneous. This is often taken as interference in the transmission of a message. Therefore, it is important to consider when sending a "dot", for example, the probability that a "dot" was received. This is represented by: ${\displaystyle P({\text{dot sent }}|{\text{ dot received}})=P({\text{dot received }}|{\text{ dot sent}}){\frac {P({\text{dot sent}})}{P({\text{dot received}})}}.}$ In Morse code, the ratio of dots to dashes is 3:4 at the point of sending, so the probabilities of a "dot" and "dash" are ${\displaystyle P({\text{dot sent}})={\frac {3}{7}}{\text{ and }}P({\text{dash sent}})={\frac {4}{7}}}$. If it is assumed that the probability that a dot is transmitted as a dash is 1/10, and that the probability that a dash is transmitted as a dot is likewise 1/10, then Bayes's rule can be used to calculate ${\displaystyle P({\text{dot received}})}$.

${\displaystyle P({\text{dot received}})=P({\text{dot received}}\cap {\text{dot sent}})+P({\text{dot received}}\cap {\text{dash sent}})}$

${\displaystyle P({\text{dot received}})=P({\text{dot received}}\mid {\text{dot sent}})P({\text{dot sent}})+P({\text{dot received}}\mid {\text{dash sent}})P({\text{dash sent}})}$

${\displaystyle P({\text{dot received}})={\frac {9}{10}}\times {\frac {3}{7}}+{\frac {1}{10}}\times {\frac {4}{7}}={\frac {31}{70}}}$

Now, ${\displaystyle P({\text{dot sent}}\mid {\text{dot received}})}$ can be calculated:

${\displaystyle P({\text{dot sent}}\mid {\text{dot received}})=P({\text{dot received}}\mid {\text{dot sent}}){\frac {P({\text{dot sent}})}{P({\text{dot received}})}}={\frac {9}{10}}\times {\frac {\frac {3}{7}}{\frac {31}{70}}}={\frac {27}{31}}}$^[14]

Dependent Events (Six-Sided Die)

Consider the experiment of rolling a fair six-sided die, where the sample space is ${\displaystyle S=\{1,2,3,4,5,6\}}$. Let event A denote the occurrence of an even number, so ${\displaystyle A=\{2,4,6\}}$, and let event B denote the occurrence of a number less than or equal to 3, so ${\displaystyle B=\{1,2,3\}}$. Before any additional information is available, the probabilities of these events are ${\displaystyle P(A)={\frac {3}{6}}={\frac {1}{2}}}$ and ${\displaystyle P(B)={\frac {3}{6}}={\frac {1}{2}}}$. The intersection of the two events is ${\displaystyle A\cap B=\{2\}}$, and therefore ${\displaystyle P(A\cap B)={\frac {1}{6}}}$. If the events were statistically independent, we would expect ${\displaystyle P(A\cap B)=P(A)P(B)}$. However, ${\displaystyle {\frac {1}{6}}\neq {\frac {1}{2}}\times {\frac {1}{2}}={\frac {1}{4}}}$, indicating that events A and B are not independent but are instead statistically dependent.

Now suppose that an observer reveals that event A has occurred; that is, the outcome is known to be an even number. This new information reduces the effective sample space from ${\displaystyle S=\{1,2,3,4,5,6\}}$ to ${\displaystyle A=\{2,4,6\}}$. Within this reduced sample space, event B can occur only when the outcome is 2, so the relevant event becomes ${\displaystyle A\cap B=\{2\}}$. Consequently, the conditional probability of event B given that event A has occurred is ${\displaystyle P(B\mid A)={\frac {P(A\cap B)}{P(A)}}={\frac {1/6}{1/2}}={\frac {1}{3}}}$. Thus, the probability of event B changes from ${\displaystyle {\frac {1}{2}}}$ to ${\displaystyle {\frac {1}{3}}}$ after the additional information is provided. This change in probability demonstrates that knowledge of event A affects the likelihood of event B, confirming that the two events are statistically dependent. In such situations, additional information reduces uncertainty and can significantly influence rational decision-making, prediction, and betting behaviour.

Independent Events (Eight-Sided Die)

Consider the experiment of rolling a fair eight-sided die with sample space ${\displaystyle S=\{1,2,3,4,5,6,7,8\}}$. Let event A represent the outcome being less than 5, so that ${\displaystyle A=\{1,2,3,4\}}$, and let event B represent the outcome being either 3, 4, 5, or 6, so that ${\displaystyle B=\{3,4,5,6\}}$. Initially, the probabilities of these events are ${\displaystyle P(A)={\frac {4}{8}}={\frac {1}{2}}}$ and ${\displaystyle P(B)={\frac {4}{8}}={\frac {1}{2}}}$. The intersection of the two events is ${\displaystyle A\cap B=\{3,4\}}$, and therefore ${\displaystyle P(A\cap B)={\frac {2}{8}}={\frac {1}{4}}}$. Since ${\displaystyle P(A)P(B={\frac {1}{2}}\times {\frac {1}{2}}={\frac {1}{4}}}$, we obtain ${\displaystyle P(A\cap B)=P(A)P(B)}$, which confirms that events A and B are statistically independent.

Now suppose that an observer reveals that event (A) has occurred. This information reduces the effective sample space from ${\displaystyle S=\{1,2,3,4,5,6,7,8\}}$ to ${\displaystyle A=\{1,2,3,4\}}$. Within this reduced sample space, the outcomes that also satisfy event B are ${\displaystyle A\cap B=\{3,4\}}$. Consequently, the conditional probability of event B given that event A has occurred is ${\displaystyle P(B\mid A)={\frac {P(A\cap B)}{P(A)}}={\frac {1/4}{1/2}}={\frac {1}{2}}}$. Notice that this conditional probability is equal to the original probability of event B, that is, ${\displaystyle P(B\mid A)=P(B)}$. Therefore, learning that event A has occurred does not change the likelihood of event B. The information provided by event A has no effect on the uncertainty associated with event B, demonstrating the fundamental property of statistical independence: knowledge of one event does not alter the probability of the other event.

Interpretation

Statistical independence can therefore be interpreted as the absence of informational value between events. If knowledge of one event changes the probability of another event, the events are dependent. Conversely, if the probability remains unchanged after obtaining new information, the events are independent.

Formally, two events A and B are independent if and only if:

${\displaystyle P(B\mid A)=P(B),}$ which is equivalent to: ${\displaystyle P(A\cap B)=P(A)P(B).}$

In this sense, independence implies that observing one event provides no additional information about the occurrence of the other event.

En general, no se puede asumir que P ( A | B )  ≈ P ( B | A ). Esto puede ser un error insidioso, incluso para aquellos que están muy familiarizados con la estadística. ^{[ 17 ]} La relación entre P ( A | B ) y P ( B | A ) viene dada por el teorema de Bayes : 

${\displaystyle {\begin{aligned}P(B\mid A)&={\frac {P(A\mid B)P(B)}{P(A)}}\\\Leftrightarrow {\frac {P(B\mid A)}{P(A\mid B)}}&={\frac {P(B)}{P(A)}}\end{aligned}}}$

Es decir, P ( A | B )  ≈ P ( B | A ) solo si P ( B )/ P ( A ) ≈ 1, o equivalentemente, P ( A ) ≈ P ( B ).     

En general, no se puede asumir que P ( A )  ≈ P ( A | B ). Estas probabilidades están vinculadas a través de la ley de la probabilidad total : 

${\displaystyle P(A)=\sum _{n}P(A\cap B_{n})=\sum _{n}P(A\mid B_{n})P(B_{n}).}$

donde los eventos${\displaystyle (B_{n})}$formar una partición contable de${\displaystyle \Omega }$.

Esta falacia puede surgir a través de un sesgo de selección . ^{[ 18 ]} Por ejemplo, en el contexto de una reclamación médica, sea S _C el evento de que una secuela (enfermedad crónica) S ocurra como consecuencia de una circunstancia (afección aguda) C. Sea H el evento de que un individuo busque ayuda médica. Supongamos que en la mayoría de los casos, C no causa S (de modo que P ( S _C ) es baja). Supongamos también que la atención médica solo se busca si S ha ocurrido debido a C. Por lo tanto, a partir de la experiencia de los pacientes, un médico puede concluir erróneamente que P ( S _C ) es alta. La probabilidad real observada por el médico es P ( S _C | H ).

No tener en cuenta la probabilidad previa, ya sea parcial o totalmente, se denomina negligencia de la tasa base . Lo contrario, un ajuste insuficiente a partir de la probabilidad previa, se denomina conservadurismo .