広島大学 先進理工系科学研究科 電気システム制御プログラム 2022年8月実施 専門科目I A-3
Author
Miyake , 祭音Myyura
Description
事象 A , B , C A, B, C A , B , C は独立で P ( A ) = 1 3 , P ( A ∩ B ) = 1 5 , P ( A ∪ C ) = 3 7 P(A) = \frac{1}{3}, P(A \cap B) = \frac{1}{5}, P(A \cup C) = \frac{3}{7} P ( A ) = 3 1 , P ( A ∩ B ) = 5 1 , P ( A ∪ C ) = 7 3 を満たすとする。ただし、P ( D ) P(D) P ( D ) は事象 D D D の確率を表す。このとき、確率 P ( B ) , P ( C ) P(B), P(C) P ( B ) , P ( C ) および条件付き確率 P ( A ∪ B ∣ C ) P(A \cup B | C) P ( A ∪ B ∣ C ) を求めよ。
Suppose that independent events A , B A, B A , B and C C C satisfy P ( A ) = 1 3 , P ( A ∩ B ) = 1 5 P(A) = \frac{1}{3}, P(A \cap B) = \frac{1}{5} P ( A ) = 3 1 , P ( A ∩ B ) = 5 1 and P ( A ∪ C ) = 3 7 P(A \cup C) = \frac{3}{7} P ( A ∪ C ) = 7 3 , where P ( D ) P(D) P ( D ) stands for the probability of an event D D D . Find the probabilities P ( B ) P(B) P ( B ) , P ( C ) P(C) P ( C ) and the conditional probability P ( A ∪ B ∣ C ) P(A \cup B | C) P ( A ∪ B ∣ C ) .
確率変数 X X X の確率密度関数が f ( x ) = exp { − ( a x 2 + b x + c ) } f(x) = \exp\{-(ax^2 + bx + c)\} f ( x ) = exp { − ( a x 2 + b x + c )} で与えられている。ただし、a , b , c a, b, c a , b , c は実数で a > 0 a > 0 a > 0 とする。
(1). X X X の期待値 E ( X ) E(X) E ( X ) と分散 V ( X ) V(X) V ( X ) をそれぞれ a , b a, b a , b を用いて表せ。
(2). E ( X ) = 1 , V ( X ) = 3 E(X) = 1, V(X) = 3 E ( X ) = 1 , V ( X ) = 3 のとき c c c の値を求めよ。
Suppose that a random variable X X X has the probability density function f ( x ) = exp { − ( a x 2 + b x + c ) } f(x) = \exp\{-(ax^2 + bx + c)\} f ( x ) = exp { − ( a x 2 + b x + c )} , where a , b a, b a , b and c c c are real numbers with a > 0 a > 0 a > 0 .
(1). Express the expectation E ( X ) E(X) E ( X ) and the variance V ( X ) V(X) V ( X ) of X X X by using a a a and b b b .
(2). Determine the value of c c c if E ( X ) = 1 E(X) = 1 E ( X ) = 1 and V ( X ) = 3 V(X) = 3 V ( X ) = 3 .
题目描述
设事件 A , B , C A,B,C A , B , C 相互独立,且
P ( A ) = 1 3 P(A)=\frac13 P ( A ) = 3 1 、P ( A ∩ B ) = 1 5 P(A\cap B)=\frac15 P ( A ∩ B ) = 5 1 、P ( A ∪ C ) = 3 7 P(A\cup C)=\frac37 P ( A ∪ C ) = 7 3 ,其中 P ( D ) P(D) P ( D ) 表示事件 D D D 的概率。求 P ( B ) P(B) P ( B ) 、P ( C ) P(C) P ( C ) 以及条件概率 P ( A ∪ B ∣ C ) P(A\cup B\mid C) P ( A ∪ B ∣ C ) 。
随机变量 X X X 的概率密度函数为
f ( x ) = exp { − ( a x 2 + b x + c ) } f(x)=\exp\{-(ax^2+bx+c)\} f ( x ) = exp { − ( a x 2 + b x + c )} ,其中 a , b , c a,b,c a , b , c 为实数且 a > 0 a>0 a > 0 。
用 a , b a,b a , b 表示期望 E ( X ) E(X) E ( X ) 和方差 V ( X ) V(X) V ( X ) 。
当 E ( X ) = 1 E(X)=1 E ( X ) = 1 、V ( X ) = 3 V(X)=3 V ( X ) = 3 时,求 c c c 的值。
Kai
事象 A , B , C A,B,C A , B , C が独立であることから、
P ( A ∩ B ) = P ( A ) P ( B ) , P ( B ∩ C ) = P ( B ) P ( C ) , P ( C ∩ A ) = P ( C ) P ( A ) , P ( A ∩ B ∩ C ) = P ( A ) P ( B ) P ( C ) \begin{aligned}
P(A \cap B) &= P(A)P(B),
\\
P(B \cap C) &= P(B)P(C),
\\
P(C \cap A) &= P(C)P(A),
\\
P(A \cap B \cap C) &= P(A)P(B)P(C)
\end{aligned} P ( A ∩ B ) P ( B ∩ C ) P ( C ∩ A ) P ( A ∩ B ∩ C ) = P ( A ) P ( B ) , = P ( B ) P ( C ) , = P ( C ) P ( A ) , = P ( A ) P ( B ) P ( C )
が成り立つ。
(i)
P ( B ) = P ( A ∩ B ) P ( A ) = 1 5 1 3 = 3 5 \begin{aligned}
P(B)
&= \frac{P(A \cap B)}{P(A)}
\\
&= \frac{\frac{1}{5}}{\frac{1}{3}}
\\
&= \frac{3}{5}
\end{aligned} P ( B ) = P ( A ) P ( A ∩ B ) = 3 1 5 1 = 5 3
(ii)
P ( A ∪ C ) = P ( A ) + P ( C ) − P ( A ∩ C ) = P ( A ) + P ( C ) − P ( A ) P ( C ) = P ( A ) + ( 1 − P ( A ) ) P ( C ) ∴ P ( C ) = P ( A ∪ C ) − P ( A ) 1 − P ( A ) = 3 7 − 1 3 1 − 1 3 = 1 7 \begin{aligned}
P(A \cup C)
&= P(A) + P(C) - P(A \cap C)
\\
&= P(A) + P(C) - P(A)P(C)
\\
&= P(A) + (1-P(A)) P(C)
\\
\therefore \ \
P(C)
&= \frac{P(A \cup C) - P(A)}{1 - P(A)}
\\
&= \frac{\frac{3}{7} - \frac{1}{3}}{1 - \frac{1}{3}}
\\
&= \frac{1}{7}
\end{aligned} P ( A ∪ C ) ∴ P ( C ) = P ( A ) + P ( C ) − P ( A ∩ C ) = P ( A ) + P ( C ) − P ( A ) P ( C ) = P ( A ) + ( 1 − P ( A )) P ( C ) = 1 − P ( A ) P ( A ∪ C ) − P ( A ) = 1 − 3 1 7 3 − 3 1 = 7 1
(iii)
P ( ( A ∪ B ) ∩ C ) = P ( ( A ∩ C ) ∪ ( B ∩ C ) ) = P ( A ∩ C ) + P ( B ∩ C ) − P ( A ∩ B ∩ C ) = P ( A ) P ( C ) + P ( B ) P ( C ) − P ( A ) P ( B ) P ( C ) = 11 105 ∴ P ( A ∪ B ∣ C ) = P ( ( A ∪ B ) ∩ C ) P ( C ) = 11 105 1 7 = 11 15 \begin{aligned}
P((A \cup B) \cap C)
&= P((A \cap C) \cup (B \cap C))
\\
&= P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)
\\
&= P(A)P(C) + P(B)P(C) - P(A)P(B)P(C)
\\
&= \frac{11}{105}
\\
\therefore \ \
P(A \cup B \mid C)
&= \frac{P((A \cup B) \cap C)}{P(C)}
\\
&= \frac{\frac{11}{105}}{\frac{1}{7}}
\\
&= \frac{11}{15}
\end{aligned} P (( A ∪ B ) ∩ C ) ∴ P ( A ∪ B ∣ C ) = P (( A ∩ C ) ∪ ( B ∩ C )) = P ( A ∩ C ) + P ( B ∩ C ) − P ( A ∩ B ∩ C ) = P ( A ) P ( C ) + P ( B ) P ( C ) − P ( A ) P ( B ) P ( C ) = 105 11 = P ( C ) P (( A ∪ B ) ∩ C ) = 7 1 105 11 = 15 11
平方完成すると
a x 2 + b x + c = a ( x + b 2 a ) 2 + c − b 2 4 a . ax^2+bx+c
=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}. a x 2 + b x + c = a ( x + 2 a b ) 2 + c − 4 a b 2 .
したがって正規分布との比較から
E ( X ) = − b 2 a , V ( X ) = 1 2 a . \boxed{E(X)=-\frac{b}{2a}},\qquad
\boxed{V(X)=\frac{1}{2a}}. E ( X ) = − 2 a b , V ( X ) = 2 a 1 .
E ( X ) = 1 E(X)=1 E ( X ) = 1 , V ( X ) = 3 V(X)=3 V ( X ) = 3 より a = 1 / 6 a=1/6 a = 1/6 , b = − 1 / 3 b=-1/3 b = − 1/3 である。さらに
1 = ∫ − ∞ ∞ f ( x ) d x = e − c + b 2 / ( 4 a ) π a 1=\int_{-\infty}^{\infty}f(x)\,dx
=e^{-c+b^2/(4a)}\sqrt{\frac{\pi}{a}} 1 = ∫ − ∞ ∞ f ( x ) d x = e − c + b 2 / ( 4 a ) a π
なので
c = b 2 4 a + 1 2 log π a = 1 6 + 1 2 log ( 6 π ) . \boxed{c=\frac{b^2}{4a}+\frac12\log\frac{\pi}{a}
=\frac16+\frac12\log(6\pi)}. c = 4 a b 2 + 2 1 log a π = 6 1 + 2 1 log ( 6 π ) .