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On the Wikipedia page for Log-normal distribution
It is written that
$$E[X|X>k]Pr(X>k)=\int_{k}^{\infty}xf_{X}(x)dx$$
I know it is probably simple, but I am still wondering the derivation. Since I know that
$$E[X|Y=y]=\int_{\mathbb{R}}xf_{X|Y}(x|y)dx$$
But I don't know why the first integral is always true.
One way is to think about the conditional density. The density of $X|X>k$ is zero when $X\leq k$, so it's proportional to $f_X(x)I(X>k)$. The constant of proportionality is given by the fact that a density has to integrate to 1, so the conditional density is $$g_X(x)=\frac{f_X(x)I(X>k)}{P(X>k)}$$ So, $$E[X|X>k]=\int_{-\infty}^\infty x\frac{f_X(x)I(X>k)}{P(X>k)}\,dx=\int_k^\infty x\frac{f_X(x)}{P(X>k)}\,dx$$ where the last step is based on noticing that the integral up to $k$ has to be zero. This then rearranges to give what you want.
The first equation can be derived using two fundamental formulas:
The formula for the expectation of a random variable $X$ given event $A$: $$E(X\mid A)=\frac{E(XI(A))}{P(A)}\tag1$$
The formula for the expectation of a function $h$ of a random variable $X$ (often described as the law of the unconscious statistician): If $X$ has density $f_X$, then $$ E [h(X)] = \int h(x) f_X(x)\, dx.\tag2$$
In your situation, apply formula (1) with $A:=\{X>k\}$, obtaining $$E(X\mid X>k)=\frac{E [XI(X>k)]}{P(X>k)}.$$ Now apply (2), noting that $XI(X>k)$ can be written $h(X)$, where $h$ has the form $$ h(x):=\begin{cases} x&\text{if $x>k$}\\ 0&\text{if $x\le k$} \end{cases}. $$ This gives $$ E [XI(X>k)]=\int_k^\infty xf_X(x)\,dx.$$
