91Ó°ÊÓ

Uniform distribution is a sort of probability distribution in statistics in which all events are equally likely. Because the chances of drawing a heart, a club, a diamond, or a spade are equal, a deck of cards contains uniform distributions. Because the likelihood of receiving heads or tails in a coin toss is the same, a coin has a uniform distribution.

Let \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)each be an independent random variable with a uniform distribution on the interval \({\rm{(100,200)}}\).

\(\begin{array}{l}{\rm{a = 100 }}\\{\rm{b = 200}}\end{array}\)

On the interval between the borders (0 elsewhere), the probability density function of a uniform distribution is the reciprocal of the difference of the boundaries:

\(\begin{aligned}{{\rm{f}}_{\rm{X}}}(x) &= \frac{{\rm{1}}}{{{\rm{b - a}}}}\\ &= \frac{{\rm{1}}}{{{\rm{200 - 100}}}}\\ &= \frac{{\rm{1}}}{{{\rm{100}}}}\\ &= 0 {\rm{.01}}\end{aligned}\)

The integral of the probability density function is the cumulative distribution function:

\({{\rm{F}}_{\rm{X}}}{\rm{(x) = }}\int_{{\rm{100}}}^{\rm{x}} {\rm{f}} {\rm{(x)dx = }}\int_{{\rm{100}}}^{\rm{x}} {\rm{0}} {\rm{.01dx = }}\left. {{\rm{(0}}{\rm{.01x)}}} \right|_{{\rm{100}}}^{\rm{x}}{\rm{ = 0}}{\rm{.01x - 1}}\)

Such that, the probability density function is \({\rm{0}}{\rm{.01}}\).

Let \(Y = \max \left( {{X_1},{X_2}, \ldots ,{X_n}} \right).\)be the case. The probability to the left of \(y\)is the cumulative distribution function at \(y\)

\({F_Y}(y) = P(Y \le y) = P\left( {\max \left( {{X_1},{X_2}, \ldots ,{X_n}} \right) \le y} \right) = P\left( {{X_1} \le y,{X_2} \le y, \ldots ,{X_n} \le y} \right)\)

For independent events, use the following multiplication rule:

\(P(A{\rm{ and }}B) = P(A) \times P(B)\)

We may use the multiplication rule for independent events since \({X_1},{X_2}, \ldots ,{X_n}\), are independent:

\(\begin{aligned}{F_Y}(y) &= P\left( {{X_1} \le y} \right)P\left( {{X_2} \le y} \right) \ldots P\left( {{X_n} \le y} \right)\\ &= P(X \le y)P(X \le y) \ldots P(X \le y)\\ &= {(P(X \le y))^n}\\ &= {\left( {{F_X}(y)} \right)^n}\\ &= {(0.01y - 1)^n}\end{aligned}\)

The pdf is the derivative of the cumulative distribution function

\(\begin{aligned}{{\rm{f}}_{\rm{Y}}}(y) &= \frac{{\rm{d}}}{{{\rm{dy}}}}\\{{\rm{F}}_{\rm{Y}}}(y) &= \frac{{\rm{d}}}{{{\rm{dy}}}}{{\rm{(0}}{\rm{.01y - 1)}}^{\rm{n}}}\\& = 0 {\rm{.01n(0}}{\rm{.01y - 1}}{{\rm{)}}^{{\rm{n - 1}}}}\end{aligned}\)

Hence, the derivative of the cumulative distribution function is \({\rm{0}}{\rm{.01n(0}}{\rm{.01y - 1}}{{\rm{)}}^{{\rm{n - 1}}}}\)

The expected value is calculated by multiplying the integral over all possible values by the pdf:

\(E(Y) = \int_{ - \infty }^{ + \infty } y {f_Y}(y)dy = \int_{100}^{200} 0 .01ny{(0.01y - 1)^{n - 1}}dy\)

Let \(u = 0.01{\rm{ }}y - 1,{\rm{ }}y = 100\left( {u + 1} \right),{\rm{ }}d{\rm{ }}y = 100{\rm{ }}d{\rm{ }}u\)

\(\begin{aligned}Let u &= 0 {\rm{.01y - 1,}}\\ y &= 100(u + 1),\\dy &= 100du\\ &= 100(0 {\rm{.01n)}}\int_{{\rm{y = 100}}}^{{\rm{y = 200}}} {\rm{1}} {\rm{00(u + 1)}}{{\rm{u}}^{{\rm{n - 1}}}}{\rm{du}}\\ &= 100n \int_{{\rm{y = 100}}}^{{\rm{y = 200}}} {\left( {{{\rm{u}}^{\rm{n}}}{\rm{ + }}{{\rm{u}}^{{\rm{n - 1}}}}} \right)} {\rm{du}}\\ &= \left. {{\rm{100n}}\left( {\frac{{{{\rm{u}}^{{\rm{n + 1}}}}}}{{{\rm{n + 1}}}}{\rm{ + }}\frac{{{{\rm{u}}^{\rm{n}}}}}{{\rm{n}}}} \right)} \right|_{{\rm{y = 100}}}^{{\rm{y = 200}}}\;\;\;\\ &= \left. {{\rm{100n}}\left( {\frac{{{{{\rm{(0}}{\rm{.01y - 1)}}}^{{\rm{n + 1}}}}}}{{{\rm{n + 1}}}}{\rm{ + }}\frac{{{{{\rm{(0}}{\rm{.01y - 1)}}}^{\rm{n}}}}}{{\rm{n}}}} \right)} \right|_{{\rm{y = 100}}}^{{\rm{y = 200}}}\end{aligned}\)

\(\begin{aligned} &= 100n\left( {\frac{{{1^{n + 1}}}}{{n + 1}} + \frac{{{1^n}}}{n}} \right)\\ &= 100n\left( {\frac{1}{{n + 1}} + \frac{1}{n}} \right)\\ &= \frac{{100(2n + 1)}}{{n + 1}}\end{aligned}\)

As a result, he can anticipate to make:\(E(Y) = \frac{{100(2n + 1)}}{{n + 1}}\)