Multidimensional probability distribution

Exercise 7

1. If one bronze coin is thrown and the front side appears, it is expressed as 1, and if the back side appears, it is expressed as 0. Answer the following questions, setting the random variables for the appearance of each bronze coin table as $X,Y,Z$ when throwing three bronze coins.

(a)

Find the distribution function of the sum $X+Y+Z$．

(b)

Find the expectaion and the distribution of the sum $X+Y+Z$．

2. The probability density of $X$ is given by the

$$f(x) = \left\{\begin{array}{cl} 2x & 0 \leq x < 1 \\ 0 & \mbox{other} \end{array} \right.$$

，Find the probability density, mean, and variance of the random variables $Y$ and $Z$.

(a)

$Y = 2X + 3$

(b)

$Z = X^2$

3. Let the random variables $X and Y$ be the rolls of the two dice

(a)

Find the expectation of $XY$．

(b)

Find the variance of $XY$．

Answer

1.

a Let $W = X + Y + Z$. Then the domain of $W$ is $\{0,1,2,3\}$．The probabilty istribution $h(i)$ is given by $P(W = i)$. Consider $h(0)$． $Since h(0) = P(W = 0)$，It's the same as throwing three bronze coins and finding the probability that all tails come out. All tail combinations are 1 out of 8．Thus, $P(W = 0) = \frac{1}{8}$．Similarly for $W = 1,2,3$,

$$\begin{array}{l} h(0) = P(W = 0) = {3 \choose 0}(\frac{1}{2})... ...) = {3 \choose 3}(\frac{1}{2})^{3} = \frac{1}{8}\\ \end{array}$$

We next find the distribution function $H(i)$. Since $H(i) = P(W \leq i)$,

$$\begin{array}{l} H(0) = P(W \leq 0) = \frac{1}{8}\\ H(1) = P... ...8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1\\ \end{array}$$

b The expectation of $W$ is $E(W)$ and

$$E(W) = E(x + Y +Z) = E(X) + E(Y) + E(Z)$$

$$E(X) = 0\cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$$

$$E(Y) = 0\cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$$

$$E(Z) = 0\cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$$

より

$$E(W) = \frac{3}{2}$$

2.

a To find the probability density function $g(x)$ of $Y$, examine the distribution function $G(y)$ of $Y$ and use the relation of $g(y) = G'(y)$.

The distribution function $F(x)$ of $X$ is $F(x) = P(X \leq x)$, and the distribution function $G(y)$ of $Y$ is $G(y) = P(Y \leq y)$ given by

$$G(y) = P(Y \leq y) = P(2X+3 \leq y) = P(X \leq \frac{y - 3}{2}) = F(\frac{y-3}{2})$$

． Note that the probability density function of $X$ is given by $f(x) = 2x \ (0 \leq x <1)$, so the probability density function $g(y)$ of $Y$ is，

$$g(y) = G'(y) = F'(\frac{y-3}{2})\frac{1}{2} = \frac{1}{2}f(\frac{y-3}{2}) = \frac{y-3}{2}$$

The expectation of $Y$ is $E(Y) = E(2X + 3) = 2E(X) + E(3) = 2E(X) + 3$. Then to find $E(Y)$, it is enough to find $E(X)$． $${E(X) = \int_{-\infty}^{\infty}xf(x) \ dx }$$. Then

$$E(X) = \int_{0}^{1}x(2x) dx = \int_{0}^{1}2x^2 dx = \frac{2}{3}x^3\mid_{0}^{1} = \frac{2}{3}$$

Therefore， $E(Y) = E(2X + 3) = 2E(X) + 3 = \frac{4}{3} + 3 = \frac{13}{3}$

Note : Since the probability density function of $Y$ was calculated, directly calculate $E(Y)$ by $E(Y) = \int_{0}^{1} \frac{y-3}{2}dy$.

Finally, we find the variance of $Y$．Note that $V(2X+3) = V(2X),V(2X) = 2^{2}V(X),V(X) = E(X^2) - (E(X))^2$. Then to find $V(Y)$, we need to find $E(X^2)$.

$$E(X^2) = \int_{-\infty}^{\infty}x^2 f(x)dx = \int_{0}^{1}2x^3 dx = \frac{1}{2}x^4 \mid_{0}^{1} = \frac{1}{2}$$

より

$$V(Y) = V(2X+3) = 4V(X) = 4[E(X^2) - (E(X))^2] = 4[\frac{1}{2} - (\frac{2}{3})^2] = \frac{2}{9}$$

b To find the probability density function $h(z)$ of $Z$, examine the distribution function $H(z)$ of $Z$ and use the relation of $h(z) = H'(z)$

The distribution function $F(x)$ of $X$ is $F(x) = P(X \leq x)$, and the distribution function $H(z)$ of $Z$ is $H(z) = P(Z \leq z)$. Since it is given by，

$$H(z) = P(Z \leq z) = P(X^2 \leq z) = P(\vert X\vert \leq \sqrt{z}) = P(X \leq \sqrt{z}) - P(X \leq -\sqrt{z}) = F(\sqrt{z}) - F(-\sqrt{z})$$

Thus the probability density function $h(z)$ of $Z$ is

$$h(z) = H'(z) = F'(\sqrt{z})(\sqrt{z})' - F'(-\sqrt{z})(-\sqrt{z})' = \frac{1}{2\sqrt{z}}(f(\sqrt{z}) + f(-\sqrt{z}))$$

Note that

$$f(x) = \left\{\begin{array}{ll} 2x & (0 \leq x < 1)\\ 0 & (other) \end{array}\right.$$

Then $f(\sqrt{z}) = 2\sqrt{z},f(-\sqrt{z}) = 0$．Therefore,， $h(z) = \frac{1}{2\sqrt{z}}2\sqrt{z} = 1$．

We next find the expectation.．

$${E(X) = \int_{-\infty}^{\infty}xf(x) \ dx }$$.

$$E(Z) = E(X^2) = \int_{-\infty}^{\infty}x^2 f(x)dx = \int_{0}^{1}2x^3 dx = \frac{1}{2}x^4 \mid_{0}^{1} = \frac{1}{2}$$

Finally, find the variance of $Z$.

$$V(Z) = V(X^2) = \int_{-\infty}^{\infty}x^2 f(x)dx = \int_{0}^{1}2x^3 dx = \frac{1}{2}x^4 \mid_{0}^{1} = \frac{1}{2}$$

implies

$$V(Y) = V(2X+3) = 4V(X) = 4[E(X^2) - (E(X))^2] = 4[\frac{1}{2} - (\frac{2}{3})^2] = \frac{2}{9}$$

3.

a Let $W = XY$. Then $X,Y$ are independent and the expectation is given by

$$E(W) = E(XY) = E(X)E(Y)$$

Since $X$ is the value of the pip that appears when one die is thrown, its expected value $E(X)$ is

$$E(X) = 1\cdot\frac{1}{6} + 2\cdot\frac{1}{6} + 3\cdot\frac{1}{6} ... ...frac{1}{6} + 5\cdot\frac{1}{6} + 6\cdot\frac{1}{6} = \frac{21}{6} = \frac{7}{2}$$

Similarly， $E(Y) = \frac{7}{2}$. Then the expectation $E(W)$ is $E(W) = (\frac{7}{2})^2$．

b Let $W = XY$. Then the varaiance is

$$V(W) = E(W^2) - (E(W))^2 = E((XY)^2) - (\frac{7}{2})^4$$

Thus, we need to find $E((XY)^2)$．Note that$X,Y$ are independent. Then

$$E((XY)^2) = E(X^2 Y^2) = E(X^2)E(Y^2)$$

Furthermore,

$$E(X^2) = 1\cdot\frac{1}{6} + 2^{2}\frac{1}{6} + 3^{2}\frac{1}{6} + 4^{2}\frac{1}{6} + 5^{2}\frac{1}{6} + 6^{2}\frac{1}{6}$$

We use the following.

$$1^2 + 2^2 + 3^2 + \cdot + n^2 = \frac{1}{6}(n(n+1)(2n+1)$$

Then

$$E(X^2) = \frac{1}{6}\frac{1}{6}(6(7)(13)) = \frac{91}{6}$$

Similarly, $E(Y^2) = \frac{91}{6}$ and $V(W)$ is

$$V(W) = E((XY)^2) - (\frac{7}{2})^4 = (\frac{91}{6})^2 - (\frac{7}{2})^4 = \frac{11515}{144}$$