Let X be a uniform random variable on the interval [0,1]. Conditioned on the event X =x, let Y be an exponential random variable with mean 1/x
I have a question that how to find the $ f \big(Y | X=x\big) $?
It is apparent that $f_{x}\big(x\big)=1$, $0 < x< 1 $ and $0$ otherwise
$$ Y = \frac{1}{x} * e^{ \big(\frac{-y}{x}\big)} $$
Appreciate the help, thanks in advance
You are told that $Y$ conditioned on $X=x$ (for any $x\in(0..1)$) is exponentially distributed with a mean of $1/x$.
Now an exponential distributed random variable with a mean of $1/\lambda$ has a probability density function of $\lambda\mathrm e^{-\lambda y}\mathbf 1_{y\geq 0}$. Not much changes when we are talking about a conditional distribution; we merely need to note where the condition is supported too.$$\begin{align}\because\quad&(Y\mid X{=}x)~\sim~\mathcal{Exp}(x)\tag{Where $0<x<1$}\\[2ex]\therefore\quad& f_{\small Y\mid X=x}(y)=x\mathrm e^{-xy}\mathbf 1_{0\leqslant y\,,\, 0<x<1}\end{align}$$
Yes. So you may next find the marginal pdf for $Y$.
$$\begin{align}f_{\small Y}(y)&=\int_\Bbb R f_{\small X}(x)~f_{\small Y\mid X{=}x}(y)~\mathrm d x\\[1ex]&=\mathbf 1_{0\leqslant y}~\int_0^1 x~\mathrm e^{-xy}~\mathrm d x\end{align}$$