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Let \(X\) and \(Y\) be continuous random variables with joint distribution function \(f(x,y).\) Let \(f_X(x)\) be the pdf of \(X.\) The conditional distribution of \(Y\) given \(X\) is \[f_{Y|X}(y|x) = \frac{f(x,y)}{f_X(x)}\]

Notice that conditional distributions on continuous random variables is computed in a similar way to conditional events. Namely, \[P(A|B) = \frac{P(A \cap B)}{P(B)}\] The various functions are replaced with the probability function, the information about \(y\) is replaced with \(A,\) and the information about \(x\) is replaced with \(B.\)

Example: Let \(X\) and \(Y\) be random variables with joint pdf \(f(x,y) = 3x\) if \(0 \leq x \leq 1\) and \(0 \leq y \leq 1,\) and \(y < x.\) What is \(P(Y > 0.1 | X = 0.25)\)?

First, we can find the conditional distribution of \(Y\) given that \(X = 0.25.\) To do so, we need the pdf of \(X,\) which can be found by integrating \(f(x,y)\) over \(y.\) For \(0 \leq x \leq 1,\) \(y < x\) gives the integral
\begin{align}
f_X(x) & = \int_0^x 3x dy \\
& = 3x\left(\left.y\right|_0^x\right) \\
& = 3x^2
\end{align}
Next, use the formula to find \(f_{Y|X}(y|x).\)
\begin{align}
f_{Y|X}(y|x) & = \frac{f(x,y)}{f_X(x)} \\
& = \frac{3x}{3x^2} \\
& = \frac{1}{x}
\end{align}
Given \(X = 0.25,\) the value of \(Y\) must be between \(0\) and \(0.25.\) The joint distribution at \(X = 0.25\) is \(f_{Y|X}(y|0.25) = \frac{1}{0.25} = 4.\) So,
\[P(Y > 0.1 | X = 0.25) = \int_{0.1}^{0.25} 4 dy = 0.6\]

Similar to the multiplication rule for probability, there is a multiplication rule for conditional distributions. \[f(x,y) = f_{Y|X}(y|x)f_X(x)\]

Example: Suppose \(X\) has pdf \(f_X(x) = 1\) for \(x \in (0, 1).\) Also, given \(X = x,\) \(Y\) has pdf \(xe^{-xy}\) for \(y > 0.\) What is the probability that \((X, Y) \in [\frac{1}{2}, 1] \times [\frac{1}{2}, 1]?\)

Let \(f(x, y)\) be the joint distribution of \(X\) and \(Y.\) The probability that \((X, Y) \in [\frac{1}{2}, 1] \times [\frac{1}{2}, 1])\) is
\[\int_{1/2}^1 \int_{1/2}^1 f(x,y) dxdy\]
We are not given a formula for \(f(x,y),\) but by the multiplication rule we can write
\[f(x,y) = f_{Y|X}(y|x)f_X(x)\]
We are given that \(X\) is has pdf \(f_X(x) = 1\) for \(x \in (0, 1).\) Also, we are given that \(f_{Y|X}(y|x) = xe^{-xy}.\) So,
\[f(x,y) = xe^{-xy}\]
for \(x \in (0, 1)\) and \(y > 0.\) Now you can directly compute the answer.
\begin{align}
P\left((X, Y) \in \left[\frac{1}{2}, 1\right] \times \left[\frac{1}{2}, 1\right]\right) & = \int_{1/2}^1 \int_{1/2}^1 f(x,y) dxdy \\
& = \int_{1/2}^1 \int_{1/2}^1 xe^{-xy} dydx \\
& = \int_{1/2}^1 \left. -e^{-xy}\right|_{y=1/2}^1 dx \\
& = \int_{1/2}^1 e^{-x/2}-e^{-x} dx \\
& = \left. -2e^{-x/2} + e^{-x} \right|_{x=1/2}^1 \\
& = 2e^{-1/4}+e^{-1}-3e^{-1/2} \\
& \approx .106
\end{align}

Check your understanding:

For problems \(1\) and \(2,\) let \(X\) be a random variable with pdf \(f(x) = \frac{1}{10}\) for \(0 < x < 10\) and \(Y\) is a random variable defined so that when \(X = x,\) the pdf of \(Y\) is \(f(y) = \frac{1}{x}\) for \(0 < y < x.\)

1. What is the joint pdf of \(X\) and \(Y?\)

Unanswered

2. What is \(P(2 < Y < 5)?\)

Unanswered

For problems \(3\) and \(4,\) let \(U\) and \(V\) be random variables with joint pdf \(f(u,v) = \frac{u+v}{9}\) for \(0 \leq u \leq 3\), \(0 \leq v \leq 3,\) and \(u + v < 3.\)

3. What is the cdf of \(U?\)

Unanswered

4. Given \(U = 2,\) what is \(P(V < 0.25)?\)

Unanswered