39.3. Exercises

Exercise 39.1 (Random and colorblind)

The gene responsible for color blindness is located on the X chromosome. In other words, red-green color blindness is an X-linked recessive condition and is much more common in males (with only one X chromosome). According to the Howard Hughes medical institute about 7% of men and 0.4% of women are red-green colorblind. Furthermore, sccording to SCB, the Swedish population is 50,3% male and 49,7% female. What is the probability that a person selcted at random is colorblind?

Exercise 39.2 (Conditional discrete probability mass function)

The joint probability mass function of the discrete variables \(X,Y\) is

\[ p(x,y) = \frac{x+y}{18}, \quad \text{for } x,y \in \{0,1,2\}. \]

• Find the conditional probability mass function \(\pdf{y}{x}\).

• Verify that it is properly normalized.

Exercise 39.3 (Conditional probability for continuous variables)

The continuous random variables \(X,Y\) have the joint density

\[ p(x,y) = e^{-x}, \quad \text{for } 0 < y < x < \infty. \]

Find the probability \(\cprob{Y<2}{X=5}\).

Exercise 39.4 (Conditional expectation)

Assume that the continuous random variables \(X,Y\) have the joint density

\[ \p{x,y} = \frac{2}{xy}, \quad \text{for } 1 < y < x < e. \]

Find the conditional expectation \(\expect{Y \vert X=x}\).

Solutions

Solution to Exercise 39.1 (Random and colorblind)

Let \(C\), \(M\), \(F\) denote the events that a random person is colorblind, male, and female, respectively. By the law of total probability

\[\begin{align*} \prob{(C)} &= \cprob{C}{M}\prob{(M)} + \cprob{C}{F}\prob{(F)} \\ &(0.07)(0.503) + (0.004)(0.497) = 0.037. \end{align*}\]

Solution to Exercise 39.2 (Conditional discrete probability mass function)

The conditional probability mass function can be obtained from the ratio

\[ \pdf{y}{x} = \frac{p(x,y)}{p(x)}. \]

Let us therefore find the marginal probability mass function

\[ p(x) = \sum_{y=0}^2 p(x,y) = \frac{x}{18} + \frac{x+1}{18} + \frac{x+2}{18} = \frac{x+1}{6}. \]

Thus we get \(\pdf{y}{x} = \frac{(x+y)/18}{(x+1)/6} = \frac{x+y}{3(x+1)}\) for \(y \in \{0,1,2\}\).

We find that this pdf (over \(y\)) is properly normalized since

\[ \sum_{y=0}^2 \pdf{y}{x} = \frac{x+0+x+1+x+2}{3(x+1)} = 1. \]

Solution to Exercise 39.3 (Conditional probability for continuous variables)

The desired probability is

\[ \cprob{Y<2}{X=5} = \int_0^2 p_{Y|X}(y \vert 5) dy. \]

To find the conditional density \(p_{Y|X}(y \vert x)\) we need the marginal one

\[ \p{x} = \int_0^\infty \p{x,y} dy = \int_0^x e^{-x} dy = x e^{-x}, \]

for \(x > 0\). This gives

\[ p_{Y|X}(y \vert x) = \frac{\p{x,y}}{\p{x}} = \frac{e^{-x}}{x e^{-x}} \frac{1}{x}, \]

for \(0 < y < x\). Note that this is a uniform distribution for \(y\) given \(x\). Therefore

\[ \cprob{Y<2}{X=5} = \int_0^2 \frac{1}{5} dy = \frac{2}{5}. \]

Solution to Exercise 39.4 (Conditional expectation)

We need the marginal density

\[ \p{x} = \int_1^x \frac{2}{xy} dy = \frac{2 \ln x}{x}, \quad \text{for } 1 < x < e, \]

to get the conditional one

\[ p_{Y|X}(y|x) = \frac{2/xy}{2 \ln x / x} = \frac{1}{y\ln x}, \quad \text{for } 1 < y < x. \]

The conditioned expectation is therefore

\[ \expect{Y \vert X=x} = \int_1^x y p_{Y|X}(y|x) dy = \int_1^x \frac{y}{y\ln x} dy = \frac{x-1}{\ln x}. \]