<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}

\newcommand{\ds}{\displaystyle}
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{\bf Question}

Suppose that the joint pdf of $X$ and $Y$ is given by

$f(x,y)=2x\ \ 0 \leq x \leq 1,\ \ 0 \leq y \leq 1$.

\begin{description}
\item[(a)]
Find the conditional distribution of $Y$ given that $X=x$,

\item[(b)]
Find $P(Y \leq \frac{1}{2}|X=0.5)$ and $P(Y \leq
\frac{1}{2}|X=0.75)$.

\item[(c)]
Explain your answers from part $b$.
\end{description}

\medskip

{\bf Answer}

$f(x,y)=2x,\ \ 0  \leq  x  \leq  1;\ \  0  \leq  y  \leq  1 $

\begin{description}
\item[(a)]
$f_X(x)=\ds\int_0^1 2x \,dy = 2x \int_0^1 \,dy = 2x,\ \ 0 \leq x
\leq 1$

$f_Y(y)=\ds\int_0^1 2x \,dx = \left.x^2\right|_0^1 = 1,\ \ 0 \leq
y \leq 1$

Therefore $f(y|x)=\ds\frac{f(x,y)}{f_X(x)}=\ds\frac{2x}{2x}=1,\ \
0 \leq y \leq 1$


\item[(b)]
$P(Y \leq \frac{1}{2}|X=0.5)=\ds\int_0^{\frac{1}{2}} \,dy =
\frac{1}{2}=\ds\int_0^{\frac{1}{2}} f(y|x) \,dy$

$P(Y \leq \frac{1}{2}|X=0.75)=\ds\int_0^{\frac{1}{2}} f(y|X=x)
\,dy = \ds\int_0^{\frac{1}{2}} \,dy = \ds\frac{1}{2}$

\item[(c)]
The two probabilities are equal because the distribution of
$Y|X=x$ does not depend on $x$ since $X$ and $Y$ are independent.
\end{description}

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