# Conditional Distribution from Conditional density

$X$ is a random variable with $f_X (x) = 3x^2$ on the interval $(0, 1)$ and $Y$ is a random variable with conditional density $f_{Y\mid X}(y\mid x) = \frac 1 x$ when $0 < y < x$ and $0 < x < 1$.

Describe the conditional distribution of $Y$ given $X = x$, and without calculation determine $E(Y \mid X = x)$.

Update:

I have since worked out that for a fixed x , the conditional density of y given x is constant and the distribution is therefore a Uniform Distribution. I still do not know how to go about the $E(Y \mid X = x)$ without calculation though!

• What's the mean of $U(a, b)$ distribution? – math_noob Nov 15 '15 at 12:06
• @math_noob $\frac{1}2 (a+b)$ but what is (a ,b) in this case? I am confused – alessia123 Nov 15 '15 at 12:18
• $Y|X = x \sim U(0, x)$ – math_noob Nov 15 '15 at 12:28

$Y|X=x$ is a uniformly distributed random variable on the interval $(0,x)$
Hence, $$E[Y|X=x] = \int_{0}^{x} y\frac{1}{x} dy$$
$$= y^2\frac{1}{2x} |_{0}^{x}$$
$$= x^2\frac{1}{2x}$$
$$= \frac{x}{2}$$
Suppose a random variable $W$ is uniform on $[0,z]$. Then its mean would be $\frac12z$. Now, as you just stated, $Y|X$ is uniform on $[0,x]$. So then the mean is $\frac12x$.