Question 1: This article
Question 2:
https://blog.yexca.net/en/archives/201
Question 3:
https://blog.yexca.net/en/archives/200
Question 4:
https://blog.yexca.net/en/archives/202
Question 5:
https://blog.yexca.net/en/archives/203
Preface
This is my first time writing an article in a non-native language. Since I didn’t refer to other materials while writing, there might be some awkward phrasing. Also, I solved these problems myself and didn’t have a reference key, so the answers might not be correct.
“Science Tokyo” is the new name for the Tokyo Institute of Technology. I used this since I’m not sure what the official abbreviation is yet.
Copyright for the questions belongs to the Institute of Science Tokyo. They are quoted here for reference only and not for commercial use.
1
Find the following limits.
a) $\lim_{x\to\infty}\{ \log_e(2x+3)-\log_e(x) \}$
b) $\lim_{x\to 0}\frac{1-\cos x}{x\sin x}$
c) $\lim_{x\to 0}\frac{e^{3x}-\cos x}{x}$
Solution
a
$$ a=\lim_{x \to \infty}\log_e\frac{2x+3}{x}=\lim_{x\to \infty}\log_e(2+\frac{3}{x})=\log_e2 $$b
$$ \because \text{when } x \to 0, \space 1-\cos x \sim \frac{1}{2}x^2 \space \text{and} \space \sin x \sim x \\ \therefore \lim_{x\to 0}\frac{1-\cos x}{x\sin x}=\lim_{x\to 0}\frac{\frac{1}{2}x^2}{x^2}=\frac{1}{2} $$c Using L’Hôpital’s rule
$$ \lim_{x\to 0}\frac{e^{3x}-\cos x}{x}=\lim_{x\to 0}\frac{3e^{3x}+\sin x}{1}=3 $$2
Calculate the determinant of the product of the following real matrices.
$$ \begin{pmatrix} 1 & 0 & 0 \\ x & 2 & 1 \\ x^2 & 3 & 2 \end{pmatrix} \begin{pmatrix} 31 & 23 & 17 \\ 0 & 11 & 11 \\ 0 & 4 & 5 \end{pmatrix} $$The determinant of the product of two matrices is the product of their determinants.
For A:
$$ A=\begin{pmatrix} 1 & 0 & 0 \\ x & 2 & 1 \\ x^2 & 3 & 2 \end{pmatrix} $$The determinant of A:
$$ \det(A)=(-1)^{1+1} \times 1 \times \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} -0+0 =1 $$For B:
$$ B=\begin{pmatrix} 31 & 23 & 17 \\ 0 & 11 & 11 \\ 0 & 4 & 5 \end{pmatrix} $$The determinant of B:
$$ \det(B)=(-1)^{1+1} \times 31 \times \begin{vmatrix} 11 & 11 \\ 4 & 5 \end{vmatrix} -0+0 = 341 $$Therefore:
$$ \det=\det(A) \times \det(B)=1 \times 341 = 341 $$3
Find the variance $V(X)$ and the cumulative distribution function $F_X(x)$ for a continuous random variable $X$ following the probability density function $f_X(x)$ below. Assume $\lim_{x \to +0}x^n\log_e(x)=0$ (where $n$ is an integer $\ge 1$).
$$ f_X(x)=\left \{ \begin{matrix} -4x\log_e(x) & 0 \lt x \le 1 \\ 0 & x \le 0 \space \text{or} \space x \gt 1 \end{matrix} \right . $$First, consider the cumulative distribution function:
- $x \le 0$
- $x \gt 1$
- $0 \lt x \le 1$
Using integration by parts:
$$ \begin{matrix} u = \ln t & dv=4tdt \\ du = \frac{1}{t}dt & v=2t^2 \end{matrix} $$Therefore:
$$ \begin{align} F_X(x) &=-( [2t^2\ln t]_0^x - \int_0^x 2t^2\frac{1}{t}dt ) \\ &=-( 2x^2\ln x - \int_0^x 2t dt ) \\ &=-( 2x^2\ln x - t^2\mid_0^x ) \\ &=x^2 - 2x^2\ln x \end{align} $$Thus:
$$ F_X(x)= \left \{ \begin{matrix} 0 & x \le 0 \\ x^2-2x^2\ln x & 0 \lt x \le 1 \\ 1 & x \gt 1 \end{matrix} \right . $$Next, calculate the variance. The definition of variance is:
$$ V(X)=E[X^2] - (E[X])^2 $$Calculating the expected value $E[X]$:
$$ \begin{align} E[X] &= \int_{-\infty}^{\infty}xf_X(x)dx \\ &= \int_0^1x(-4x\ln x)dx \\ &= -\int_0^1 4x^2\ln xdx \end{align} $$Using integration by parts:
$$ \begin{matrix} u=\ln x & dv=4x^2dx \\ du=\frac{1}{x} & v=\frac{4}{3}x^3 \end{matrix} $$Therefore:
$$ \begin{align} E[X] &= -([\frac{4}{3}x^3\ln x]_0^1-\int_0^1\frac{4}{3}x^3\frac{1}{x}dx) \\ &= -(-\frac{4}{9}x^3\mid_0^1) \\ &= \frac{4}{9} \end{align} $$Calculating $E[X^2]$:
$$ \begin{align} E[X^2] &= \int_{-\infty}^{\infty}x^2f_X(x)dx \\ &= \int_0^1x^2(-4x\ln x)dx \\ &= -4\int_0^1x^3\ln xdx \end{align} $$Using integration by parts:
$$ \begin{matrix} u=\ln x & dv=x^3dx \\ du=\frac{1}{x} & v=\frac{1}{4}x^4 \end{matrix} $$Therefore:
$$ \begin{align} E[X^2] &= -4 \times ([\frac{1}{4}x^4\ln x]_0^1-\int_0^1\frac{1}{4}x^3dx) \\ &= -4 \times (-\frac{1}{16}x^4\mid_0^1) \\ &= \frac{1}{4} \end{align} $$So:
$$ \begin{align} V(X) &= E[X^2] - (E[X])^2 \\ &= \frac{1}{4} - (\frac{4}{9})^2 \\ &= \frac{17}{324} \end{align} $$4
In a certain manufacturing line, products are defective with a probability of 1/1000. There is an inspection method that correctly identifies a defective product with a probability of 99/100, and correctly identifies a non-defective product with a probability of 4/5. Find the probability that a product identified as defective by this method is actually defective.
Probability of being defective: $P(A)=\frac{1}{1000}$, probability of being judged as defective: $P(B)$.
From the problem:
$$ \begin{matrix} P(\bar{A})=\frac{999}{1000} & P(B\mid A)=\frac{99}{100} & P(B\mid \bar{A})=\frac{1}{5} \end{matrix} $$Therefore:
$$ P(B)=P(A)P(B\mid A)+P(\bar{A})P(B\mid \bar{A})=\frac{20079}{100000} $$Using Bayes’ theorem:
$$ P(A\mid B)=\frac{P(AB)}{P(B)}=\frac{P(A)P(B\mid A)}{P(B)}=\frac{99}{20079} $$Thus, the probability of the product being defective is $\frac{99}{20079}$.
5
In a casino, there is a roulette wheel with four pockets: A, B, C, and D. The casino claims that the probability of the ball landing in each pocket is equal. In 5 trials, the ball landed in pocket A four times. Test the hypothesis “this roulette wheel is biased toward pocket A” at a 5% significance level. Clearly state the null hypothesis $H_0$ and the alternative hypothesis $H_1$ in your answer.
$H_0$: The probability of the ball landing in pocket A is the same as the other pockets. That is:
$$ P(A)=\frac{1}{4} $$$H_1$: The probability of the ball landing in pocket A is higher than the other pockets. That is:
$$ P(A)\gt \frac{1}{4} $$The probability of the ball landing in pocket A exactly 4 times is:
$$ P(X=4)=\binom{5}{4} (\frac{1}{4})^4 \times \frac{3}{4}=\frac{15}{1024} $$The probability of the ball landing in pocket A exactly 5 times is:
$$ P(X=5)=\binom{5}{5} (\frac{1}{4})^5=\frac{1}{1024} $$Thus:
$$ P(X\ge 4)=\frac{15}{1024}+\frac{1}{1024}=\frac{1}{64} $$Since this probability ($1/64 \approx 0.0156$) is smaller than the significance level (0.05), we reject the null hypothesis. Therefore, the hypothesis that “this roulette wheel is biased toward pocket A” is statistically significant.