Why $\mathbb P(X\boldsymbol 1_{\{X>a\}}>y)=\mathbb P(X>y\mid X>a)$? I saw here in the answer of user657324 that $$\mathbb P(X\boldsymbol 1_{\{X>a\}}>y)=\mathbb P(X>y\mid X>a),\tag{*}$$
but I don't really understand why it's true. I try to ask directely to user657324, but it looks that I can't leave any comment. So why $(*)$ is true ? And in general, do we have that $$\mathbb P(XY>z)=\mathbb P(X>z\mid Y)=\mathbb E[\boldsymbol 1_{\{X>z\}}\mid Y]\ \ ?$$
And if yes, why ? 
 A: it seem not correct.
$$\mathbb P(XY>z)=\mathbb P(A)=\mathbb E(1_A)=\mathbb E(\mathbb E(1_A|Y))=\mathbb E(\mathbb P(A|Y))=\mathbb E(\mathbb P(XY>z|Y))$$
and
$$\mathbb E(\mathbb P(XY>z|Y=y))=\mathbb E(\mathbb P(Xy>z|Y=y))$$
and it is depend on $y<0$ or $y\geq 0$. 
if $Y=1_{X>a}$ so $Y\sim Ber(p=P(X>a))$
$$\mathbb P(XY>z)=\mathbb P(A)=\mathbb E(1_A)=\mathbb E(\mathbb E(1_A|Y))=
\mathbb E(g(Y))$$
$$=P(Y=1) * g(1) +P(Y=0) * g(0)$$
$$=P(Y=1) * E(1_A|Y=1)+P(Y=0)*E(1_A|Y=0)$$
$$=P(Y=1) * P(A|Y=1)+P(Y=0)*P(A|Y=0)$$
$$=P(1_{X>a}=1) * P(X1_{X>a}>z|1_{X>a}=1)+P(1_{X>a}=0)*P(X1_{X>a}>z|1_{X>a}=0)$$
$$=P(1_{X>a}=1) * P(X>z|1_{X>a}=1)+P(1_{X>a}=0)*P(0>z|1_{X>a}=0)$$
note $P(0>z|1_{X>a}=0)$ is alwayes is 1 or zero 
$$=P(X>a) * P(X>z|X>a)+P(X\leq a)*P(0>z|X\leq a)$$
for example if $z\geq 0$
$$=P(X>a) * P(X>z|X>a)+0=P(X>z , X>a)=P(X>max(a,z))$$
but if $z<0$
$$=P(X>a) * P(X>z|X>a)+P(X\leq a)=P(X>max(a,z))+P(X\leq a)$$
A: It looks wrong to me. If $y>0$, the event $X\boldsymbol 1_{\{X>a\}}>y$ is equivalent to $X>a \cap X>y$, or equivalently  $X>\max(a,y)$.
More in detail, calling $Z = 1_{\{X>a\}}$ :
$$\begin{align}
P(X Z >y) &= P(X Z >y , Z=0) + P(X Z >y, Z=1) \\
&= P(X Z >y \mid Z=0) P(Z=0) +  P(X Z >y \mid Z=1) P(Z=1)  \\
&= 0 P(Z=0) +P(X >y \mid Z=1) P(Z=1) \\
&= P(X > y \mid X>a) P(X>a)\\
&= P(X > y , X>a) \\
&= P(X > \max(y,a)) 
\end{align}
$$
BTW: the linked answer gets the correct result in spite of this error, because it has another error (the first equality) and they cancel out. Actually $E[X\mid X>a]=\mathbb E[X\cdot \boldsymbol 1_{\{X>a\}}]$.
