I need some help with this probrem:
If $X$ is a random variable with density
$f(x) =\left\{ \begin{array}{lcc} \\ c\frac{1}{x^2} & , & 1\leq X\leq2 \\ \\ 0 & , & \text{otherwise} \end{array}\right.$
If $A=X>1.5$, find the conditional density of $X$ knowing that $A$ has occurred.
I tried to solve it by defining a indicator varible:
$$I_A =\left\{ \begin{array}{lcc} \\ 1 & , & X>1.5 \\ \\ 0 & , & \text{otherwise} \end{array}\right.$$
Then I used the formula for conditional continous distribution: $$f_{X|I_A}(X|I_A)=\frac{f(X,I_A)}{f(I_a)}$$ I don't know how to find $f(X,I_A)$ since if I'm not mistaken they are not independet so I can't just muliply the marginals. How do I fin the conditional density?
First we compute $c$: $$ 1=\int_0^1 f(x)\ \mathsf dx = \int_0^1 \frac c{x^2}\ \mathsf dx = \frac12 c\implies c=2. $$
The conditional density is given by $$ f_{X\mid A}(x) = \frac{f(x)}{\mathbb P(A)}\mathsf 1_A. $$ We compute $$ \mathbb P(A) = \int_{\frac32}^2 \frac2{x^2}\ \mathsf dx = \frac13. $$ Hence, $$ f_{X\mid A}(x) = \frac2{3x^2}\mathsf 1_{\left(\frac32,2\right)}(x). $$