Showing $\mathbb{E}[X\mid X>b]=\frac{1}{\mathbb{P}(X>b)}\int_b^{\infty}xf(x)dx$

Question

Let $X$ be a continous random variable with pdf $f$ and let $b$ be a real number. Show that $$\mathbb{E}[X \mid X>b]=\frac{1}{\mathbb{P}(X>b)}\int_b^{\infty}xf(x)dx.\tag1$$

I do not know the type of distribution of $X$ but the range is $[b,\infty)$. According to definition we have that:

Let $X$ be a continous random variable and $B$ and event with $\mathbb{P}(B)>0.$ The conditional expectation of $X$ given $B$ is defined as $$\mathbb{E}[X \mid B]=\int_{-\infty}^{\infty}xf_X(x \mid B) \ dx.\tag2$$

In my case, $B=X>b$. Substituting into $(2)$ gives

$$\mathbb{E}[X \mid X>b]=\int_{b}^{\infty}xf_X(x \mid X>b) \ dx.$$

I do not get anywhere from here.

Answer 1

$\def\d{\mathrm{d}}$Because$$ P(X \leqslant x \mid X > b) = \frac{P(b < X \leqslant x)}{P(X > b)} = \frac{1}{P(X > b)} \int_b^x f_X(y) \,\d y, \quad \forall x > b $$ then$$ f_X(x \mid X > b) = \frac{f_X(x)}{P(X > b)}. \quad \forall x > b $$