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David Smith
David Smith
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There are two index sets $I= \{1, 2, \dots, m\}$ and $J = \{1, 2, \dots, n\}$. Then, I have independent random variables $X_{ij}, \forall i \in I, j \in J$. Fix $i \in I$; then we have $X_{ij}$ is identical for all $j \in J$.

Fix $k \in J$, I am wondering if I could have the following probability inequality$$ P\left(X_{1k} \leq Y, X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \leq $$$$ P\left(X_{1k} \leq Y, X_{2k} \leq Y \mid Y = \max_{I\in I} \min_{j \in J} X_{ij}\right) \leq $$ $$ P\left(X_{1k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \cdot P\left( X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right)$$$$ P\left(X_{1k} \leq Y \mid Y = \max_{I\in I} \min_{j \in J} X_{ij}\right) \cdot P\left( X_{2k} \leq Y \mid Y = \max_{I\in I} \min_{j \in J} X_{ij}\right)$$

If the inequality holds, how to show it? If the inequality does not hold, can we have some inequality of a similar format? Thank you very much!

There are two index sets $I= \{1, 2, \dots, m\}$ and $J = \{1, 2, \dots, n\}$. Then, I have independent random variables $X_{ij}, \forall i \in I, j \in J$. Fix $i \in I$; then we have $X_{ij}$ is identical for all $j \in J$.

Fix $k \in J$, I am wondering if I could have the following probability inequality$$ P\left(X_{1k} \leq Y, X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \leq $$ $$ P\left(X_{1k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \cdot P\left( X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right)$$

If the inequality holds, how to show it? If the inequality does not hold, can we have some inequality of a similar format? Thank you very much!

There are two index sets $I= \{1, 2, \dots, m\}$ and $J = \{1, 2, \dots, n\}$. Then, I have independent random variables $X_{ij}, \forall i \in I, j \in J$. Fix $i \in I$; then we have $X_{ij}$ is identical for all $j \in J$.

Fix $k \in J$, I am wondering if I could have the following probability inequality$$ P\left(X_{1k} \leq Y, X_{2k} \leq Y \mid Y = \max_{I\in I} \min_{j \in J} X_{ij}\right) \leq $$ $$ P\left(X_{1k} \leq Y \mid Y = \max_{I\in I} \min_{j \in J} X_{ij}\right) \cdot P\left( X_{2k} \leq Y \mid Y = \max_{I\in I} \min_{j \in J} X_{ij}\right)$$

If the inequality holds, how to show it? If the inequality does not hold, can we have some inequality of a similar format? Thank you very much!

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edit approved Nov 24, 2021 at 6:28
Dirk Werner
edit approved Nov 24, 2021 at 6:28
Dirk Werner
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Denote thereThere are two index sets $I= \{1, 2, \cdots, m\}$$I= \{1, 2, \dots, m\}$ and $J = \{1, 2, \cdots, n\}.$$J = \{1, 2, \dots, n\}$. Then, I have independent random variables $X_{ij}, \forall i \in I, j \in J.$$X_{ij}, \forall i \in I, j \in J$. Fix $i \in I,$$i \in I$; then we have $X_{ij}$ is identical for all $j \in J.$$j \in J$.

Fix $k \in J,$$k \in J$, I am wondering if I could have the following probability inequality$$ P\left(X_{1k} \leq Y, X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \leq P\left(X_{1k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \cdot P\left( X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right)$$$$ P\left(X_{1k} \leq Y, X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \leq $$ $$ P\left(X_{1k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \cdot P\left( X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right)$$

If the inequality holds, how to show it? If the inequality does not hold, can we have some inequality of a similar format? Thank you very much!

Denote there are two index sets $I= \{1, 2, \cdots, m\}$ and $J = \{1, 2, \cdots, n\}.$ Then, I have independent random variables $X_{ij}, \forall i \in I, j \in J.$ Fix $i \in I,$ we have $X_{ij}$ is identical for all $j \in J.$

Fix $k \in J,$ I am wondering if I could have the following probability inequality$$ P\left(X_{1k} \leq Y, X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \leq P\left(X_{1k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \cdot P\left( X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right)$$

If the inequality holds, how to show it? If the inequality does not hold, can we have some inequality of similar format? Thank you very much!

There are two index sets $I= \{1, 2, \dots, m\}$ and $J = \{1, 2, \dots, n\}$. Then, I have independent random variables $X_{ij}, \forall i \in I, j \in J$. Fix $i \in I$; then we have $X_{ij}$ is identical for all $j \in J$.

Fix $k \in J$, I am wondering if I could have the following probability inequality$$ P\left(X_{1k} \leq Y, X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \leq $$ $$ P\left(X_{1k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \cdot P\left( X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right)$$

If the inequality holds, how to show it? If the inequality does not hold, can we have some inequality of a similar format? Thank you very much!

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David Smith
David Smith
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Conditional probability inequality proof

Denote there are two index sets $I= \{1, 2, \cdots, m\}$ and $J = \{1, 2, \cdots, n\}.$ Then, I have independent random variables $X_{ij}, \forall i \in I, j \in J.$ Fix $i \in I,$ we have $X_{ij}$ is identical for all $j \in J.$

Fix $k \in J,$ I am wondering if I could have the following probability inequality$$ P\left(X_{1k} \leq Y, X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \leq P\left(X_{1k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right) \cdot P\left( X_{2k} \leq Y \mid Y \geq \max_{I\in I} \min_{j \in J} X_{ij}\right)$$

If the inequality holds, how to show it? If the inequality does not hold, can we have some inequality of similar format? Thank you very much!