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Iosif Pinelis
Iosif Pinelis
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The answer is yes. Indeed, let us write $X$ instead of $x$, according to standard notation, to distinguish between random variables (denoted by upper-case letters) and their values (denoted by lower-case letters). Let us write $P$ instead of $\Pr$, and then let us also write $A$ instead of $E$, to distinguish it from the expectation sign.

Then we need to show that \begin{equation} EX\,P(A|X)\ge P(A)\,EX \end{equation} given that \begin{equation} P(A|X>y)\ge P(A) \end{equation} for all $y$ such that $P(X>y)\ne0$.

We haveReplacing $X$ by $X-a$, we may assume that $X\ge0$. Then \begin{multline*} EX\,P(A|X)=EX\,E(1_A|X)=EX\,1_A \\ =E\int_0^\infty dy\,1_{X>y,A}=\int_0^\infty dy\,E1_{X>y,A} \\ =\int_0^\infty dy\,P(X>y,A) =\int_0^\infty dy\,P(A|X>y)P(X>y) \\ \ge\int_0^\infty dy\,P(A)P(X>y)=P(A)\int_0^\infty dy\,P(X>y)=P(A)EX, \end{multline*}\begin{multline*} EX\,P(A|X)=EX\,E(1_A|X)=EX\,1_A=E\Big(\int_0^X dy\Big)\,1_A \\ =E\Big(\int_0^\infty dy\,1_{X>y}\Big)\,1_A =E\Big(\int_0^\infty dy\,1_{X>y}1_A\Big) \\ =E\int_0^\infty dy\,1_{X>y,A}=\int_0^\infty dy\,E1_{X>y,A} \\ =\int_0^\infty dy\,P(X>y,A) =\int_0^\infty dy\,P(A|X>y)P(X>y) \\ \ge\int_0^\infty dy\,P(A)P(X>y)=P(A)\int_0^\infty dy\,P(X>y)=P(A)EX, \end{multline*} as claimed.

(The last equality follows immediately from the special case $EX=\int_0^\infty dy\,P(X>y)$ of the previously proved equality $EX\,P(A|X)=\int_0^\infty dy\,P(X>y,A)$, with $A$ replaced there by an event of probability $1$.)

The answer is yes. Indeed, let us write $X$ instead of $x$, according to standard notation, to distinguish between random variables (denoted by upper-case letters) and their values (denoted by lower-case letters). Let us write $P$ instead of $\Pr$, and then let us also write $A$ instead of $E$, to distinguish it from the expectation sign.

Then we need to show that \begin{equation} EX\,P(A|X)\ge P(A)\,EX \end{equation} given that \begin{equation} P(A|X>y)\ge P(A) \end{equation} for all $y$ such that $P(X>y)\ne0$.

We have \begin{multline*} EX\,P(A|X)=EX\,E(1_A|X)=EX\,1_A \\ =E\int_0^\infty dy\,1_{X>y,A}=\int_0^\infty dy\,E1_{X>y,A} \\ =\int_0^\infty dy\,P(X>y,A) =\int_0^\infty dy\,P(A|X>y)P(X>y) \\ \ge\int_0^\infty dy\,P(A)P(X>y)=P(A)\int_0^\infty dy\,P(X>y)=P(A)EX, \end{multline*} as claimed.

(The last equality follows immediately from the special case $EX=\int_0^\infty dy\,P(X>y)$ of the previously proved equality $EX\,P(A|X)=\int_0^\infty dy\,P(X>y,A)$, with $A$ replaced there by an event of probability $1$.)

The answer is yes. Indeed, let us write $X$ instead of $x$, according to standard notation, to distinguish between random variables (denoted by upper-case letters) and their values (denoted by lower-case letters). Let us write $P$ instead of $\Pr$, and then let us also write $A$ instead of $E$, to distinguish it from the expectation sign.

Then we need to show that \begin{equation} EX\,P(A|X)\ge P(A)\,EX \end{equation} given that \begin{equation} P(A|X>y)\ge P(A) \end{equation} for all $y$ such that $P(X>y)\ne0$.

Replacing $X$ by $X-a$, we may assume that $X\ge0$. Then \begin{multline*} EX\,P(A|X)=EX\,E(1_A|X)=EX\,1_A=E\Big(\int_0^X dy\Big)\,1_A \\ =E\Big(\int_0^\infty dy\,1_{X>y}\Big)\,1_A =E\Big(\int_0^\infty dy\,1_{X>y}1_A\Big) \\ =E\int_0^\infty dy\,1_{X>y,A}=\int_0^\infty dy\,E1_{X>y,A} \\ =\int_0^\infty dy\,P(X>y,A) =\int_0^\infty dy\,P(A|X>y)P(X>y) \\ \ge\int_0^\infty dy\,P(A)P(X>y)=P(A)\int_0^\infty dy\,P(X>y)=P(A)EX, \end{multline*} as claimed.

(The last equality follows immediately from the special case $EX=\int_0^\infty dy\,P(X>y)$ of the previously proved equality $EX\,P(A|X)=\int_0^\infty dy\,P(X>y,A)$, with $A$ replaced there by an event of probability $1$.)

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Iosif Pinelis
Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229

The answer is yes. Indeed, let us write $X$ instead of $x$, according to standard notation, to distinguish between random variables (denoted by upper-case letters) and their values (denoted by lower-case letters). Let us write $P$ instead of $\Pr$, and then let us also write $A$ instead of $E$, to distinguish it from the expectation sign.

Then we need to show that \begin{equation} EX\,P(A|X)\ge P(A)\,EX \end{equation} given that \begin{equation} P(A|X>y)\ge P(A) \end{equation} for all $y$ such that $P(X>y)\ne0$.

We have \begin{multline*} EX\,P(A|X)=EX\,E(1_A|X)=EX\,1_A \\ =E\int_0^\infty dy\,1_{X>y,A}=\int_0^\infty dy\,E1_{X>y,A} \\ =\int_0^\infty dy\,P(X>y,A) =\int_0^\infty dy\,P(A|X>y)P(X>y) \\ \ge\int_0^\infty dy\,P(A)P(X>y)=P(A)\int_0^\infty dy\,P(X>y)=P(A)EX, \end{multline*} as claimed.

(The last equality follows immediately from the special case $EX=\int_0^\infty dy\,P(X>y)$ of the previously proved equality $EX\,P(A|X)=\int_0^\infty dy\,P(X>y,A)$, with $A$ replaced there by an event of probability $1$.)

The answer is yes. Indeed, let us write $X$ instead of $x$, according to standard notation, to distinguish between random variables (denoted by upper-case letters) and their values (denoted by lower-case letters). Let us write $P$ instead of $\Pr$, and then let us also write $A$ instead of $E$, to distinguish it from the expectation sign.

Then we need to show that \begin{equation} EX\,P(A|X)\ge P(A)\,EX \end{equation} given that \begin{equation} P(A|X>y)\ge P(A) \end{equation} for all $y$ such that $P(X>y)\ne0$.

We have \begin{multline*} EX\,P(A|X)=EX\,E(1_A|X)=EX\,1_A \\ =E\int_0^\infty dy\,1_{X>y,A}=\int_0^\infty dy\,E1_{X>y,A} \\ =\int_0^\infty dy\,P(X>y,A) =\int_0^\infty dy\,P(A|X>y)P(X>y) \\ \ge\int_0^\infty dy\,P(A)P(X>y)=P(A)\int_0^\infty dy\,P(X>y)=P(A)EX, \end{multline*} as claimed.

The answer is yes. Indeed, let us write $X$ instead of $x$, according to standard notation, to distinguish between random variables (denoted by upper-case letters) and their values (denoted by lower-case letters). Let us write $P$ instead of $\Pr$, and then let us also write $A$ instead of $E$, to distinguish it from the expectation sign.

Then we need to show that \begin{equation} EX\,P(A|X)\ge P(A)\,EX \end{equation} given that \begin{equation} P(A|X>y)\ge P(A) \end{equation} for all $y$ such that $P(X>y)\ne0$.

We have \begin{multline*} EX\,P(A|X)=EX\,E(1_A|X)=EX\,1_A \\ =E\int_0^\infty dy\,1_{X>y,A}=\int_0^\infty dy\,E1_{X>y,A} \\ =\int_0^\infty dy\,P(X>y,A) =\int_0^\infty dy\,P(A|X>y)P(X>y) \\ \ge\int_0^\infty dy\,P(A)P(X>y)=P(A)\int_0^\infty dy\,P(X>y)=P(A)EX, \end{multline*} as claimed.

(The last equality follows immediately from the special case $EX=\int_0^\infty dy\,P(X>y)$ of the previously proved equality $EX\,P(A|X)=\int_0^\infty dy\,P(X>y,A)$, with $A$ replaced there by an event of probability $1$.)

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Iosif Pinelis
Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229

The answer is yes. Indeed, let us write $X$ instead of $x$, according to standard notation, to distinguish between random variables (denoted by upper-case letters) and their values (denoted by lower-case letters). Let us write $P$ instead of $\Pr$, and then let us also write $A$ instead of $E$, to distinguish it from the expectation sign.

Then we need to show that \begin{equation} EX\,P(A|X)\ge P(A)\,EX \end{equation} given that \begin{equation} P(A|X>y)\ge P(A) \end{equation} for all $y$ such that $P(X>y)\ne0$.

We have \begin{multline*} EX\,P(A|X)=EX\,E(1_A|X)=EX\,1_A \\ =E\int_0^\infty dy\,1_{X>y,A}=\int_0^\infty dy\,E1_{X>y,A} \\ =\int_0^\infty dy\,P(X>y,A) =\int_0^\infty dy\,P(A|X>y)P(X>y) \\ \ge\int_0^\infty dy\,P(A)P(X>y)=P(A)\int_0^\infty dy\,P(X>y)=P(A)EX, \end{multline*} as claimed.