## STAT501 Regression Analysis课程简介

This graduate level course offers an introduction into regression analysis. A
Credits 3 researcher is often interested in using sample data to investigate relationships, with an ultimate goal of creating a model to predict a future value for some dependent variable. The process of finding this mathematical model that best fits the data involves regression analysis.
STAT 501 is an applied linear regression course that emphasizes data analysis and interpretation. Generally, statistical regression is collection of methods for determining and using models that explain how a response variable (dependent variable) relates to one or more explanatory variables (predictor variables).

## PREREQUISITES

This graduate level course covers the following topics:

• Understanding the context for simple linear regression.
• How to evaluate simple linear regression models
• How a simple linear regression model is used to estimate and predict likely values
• Understanding the assumptions that need to be met for a simple linear regression model to be valid
• How multiple predictors can be included into a regression model
• Understanding the assumptions that need to be met when multiple predictors are included in the regression model for the model to be valid
• How a multiple linear regression model is used to estimate and predict likely values
• Understanding how categorical predictors can be included into a regression model
• How to transform data in order to deal with problems identified in the regression model
• Strategies for building regression models
• Distinguishing between outliers and influential data points and how to deal with these
• Handling problems typically encountered in regression contexts
• Alternative methods for estimating a regression line besides using ordinary least squares
• Understanding regression models in time dependent contexts
• Understanding regression models in non-linear contexts

## STAT501 Regression Analysis HELP（EXAM HELP， ONLINE TUTOR）

Exercise 1 (Should you regress $\boldsymbol{Y}$ on $\boldsymbol{X}$ or vice-versa?) The answer to that question is not a statistical question, it is a scientific one. Do you have a theory that makes one variable dependent, and the other independent? The statistical question is what difference does it make?
$$Y=\beta_0 1+\beta_1 X+\varepsilon$$
Let $\hat{\beta}_1$ be the least squares estimate of $\beta_1$ for this model.
Now suppose you try the model
$$X=\alpha_0+\alpha_1 Y+\eta .$$
Let $\hat{\alpha}_1$ be the least squares estimate of $\alpha_1$.

The question of whether to regress $Y$ on $X$ or $X$ on $Y$ depends on the scientific question you are trying to answer. If you have a theory that suggests that $X$ is the independent variable and $Y$ is the dependent variable, then it makes sense to regress $Y$ on $X$. Conversely, if your theory suggests that $Y$ is the independent variable and $X$ is the dependent variable, then you should regress $X$ on $Y$.

However, from a statistical point of view, regressing $Y$ on $X$ or $X$ on $Y$ can give different results. In general, the least squares estimates of $\beta_1$ and $\alpha_1$ are not the same. They may have the same sign and similar magnitude, but their values can be quite different.

This is because the two models are estimating different things. The model $Y=\beta_0 + \beta_1 X + \epsilon$ is estimating the effect of $X$ on $Y$, while the model $X=\alpha_0 + \alpha_1 Y + \eta$ is estimating the effect of $Y$ on $X$. In general, these effects are not the same, and therefore the estimates of $\beta_1$ and $\alpha_1$ can be different.

Furthermore, the regression of $X$ on $Y$ may be problematic if there is a high correlation between $X$ and $Y$, as this can lead to multicollinearity and unstable estimates. In this case, it may be preferable to regress $Y$ on $X$ instead.

In summary, the choice of whether to regress $Y$ on $X$ or $X$ on $Y$ depends on the scientific question you are trying to answer. From a statistical point of view, both approaches can give different results, and the choice may depend on the specific data and the presence of multicollinearity.

Exercise 2 (Anscombe’s quartet) Francis Anscombe [1] created and presented four small sets of data. You can find them on the course web site at http://www. math. caltech. edu/\%7E2015-16/2term/ma003/Data/Anscombe1.txt, Anscombe2, Anscombe3, and Anscombe4. Each data set has 11 observations on two variates labeled $X$ and $Y$. (Each file has a header line.)

1. For each data set,
(a) Compute the sample mean and standard deviation of $X$ and $Y$.
(b) Regress $Y$ on $X$ and a constant term. Explain the computations you are doing. (If you omit this information, we shall be unable to award partial credit if your calculations are incorrect.)

Sure, I can perform the requested computations for each of the four data sets in Anscombe’s quartet.

1. For Anscombe1 data set:

(a) Sample mean and standard deviation of $X$ and $Y$:

\begin{align*} \bar{X} &= 9.0, \quad \bar{Y} = 7.5 \ s_X &= 3.3166, \quad s_Y = 2.0316 \end{align*}

(b) Regression of $Y$ on $X$ and a constant term: The linear regression model is given by:

Y_i = \beta_0 + \beta_1 X_i + \epsilon_i, \qquad i=1,\ldots,11Yi​=β0​+β1​Xi​+ϵi​,i=1,…,11

where $\epsilon_i$ are the random errors with zero mean and constant variance. The least squares estimates of the regression coefficients $\beta_0$ and $\beta_1$ are given by: \begin{align*} \hat{\beta}1 &= \frac{\sum{i=1}^{11}(X_i – \bar{X})(Y_i – \bar{Y})}{\sum_{i=1}^{11}(X_i – \bar{X})^2} = 0.5 \ \hat{\beta}_0 &= \bar{Y} – \hat{\beta}_1 \bar{X} = 3 \end{align*}

Thus, the estimated regression line is:

\hat{Y} = 3 + 0.5 XY^=3+0.5X

1. For Anscombe2 data set:

(a) Sample mean and standard deviation of $X$ and $Y$:

\begin{align*} \bar{X} &= 9.0, \quad \bar{Y} = 7.5 \ s_X &= 3.3166, \quad s_Y = 2.0316 \end{align*}

(b) Regression of $Y$ on $X$ and a constant term: The linear regression model is given by:

Y_i = \beta_0 + \beta_1 X_i + \epsilon_i, \qquad i=1,\ldots,11Yi​=β0​+β1​Xi​+ϵi​,i=1,…,11

where $\epsilon_i$ are the random errors with zero mean and constant variance. The least squares estimates of the regression coefficients $\beta_0$ and $\beta_1$ are given by: \begin{align*} \hat{\beta}1 &= \frac{\sum{i=1}^{11}(X_i – \bar{X})(Y_i – \bar{Y})}{\sum_{i=1}^{11}(X_i – \bar{X})^2} = 0.5 \ \hat{\beta}_0 &= \bar{Y} – \hat{\beta}_1 \bar{X} = 3 \end{align*}

Thus, the estimated regression line is:

\hat{Y} = 3 + 0.5 XY^=3+0.5X

1. For Anscombe3 data set:

(a) Sample mean and standard deviation of $X$ and $Y$:

\begin{align*} \bar{X} &= 9.0, \quad \bar{Y} = 7.5 \ s_X &= 3.3166, \quad s_Y = 2.0316 \end{align*}

(b) Regression of $Y$ on $X$ and a constant term: The linear regression model is given by.

## Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

Statistics-lab™可以为您提供psu.edu STAT501 Regression Analysis回归分析的代写代考辅导服务！ 请认准Statistics-lab™. Statistics-lab™为您的留学生涯保驾护航。

## 数学代写|非线性优化Nonlinear optimization代考

(1) Conside the convex set $C:=\left{(x, y) \in \mathbb{R}^2 \mid x^2+y^2 \leq 1\right.$ and $\left.x, y \geq 0\right}$, the unit disc intersect the first quadrant, and let $\delta_C(x, y)$ be its indicator function. Find $\partial \delta_C\left(x_0, y_0\right)$ where $\left(x_0, y_0\right)$ is a boundary point of $C$. Hint: Draw a picture. There should be 6 cases to consider. Note that $C=C_1 \cap C_2 \cap C_3$ where $C_1=\left{(x, y) \in \mathbb{R}^2 \mid x^2+y^2 \leq 1\right}$, $C_2=\left{(x, y) \in \mathbb{R}^2 \mid-x \leq 0\right}$, and $C_3=\left{(x, y) \in \mathbb{R}^2 \mid-y \leq 0\right}$.

To find $\partial \delta_C(x_0, y_0)$, we need to find the subdifferential of the indicator function $\delta_C$ at the point $(x_0, y_0)$. Since $C$ is a convex set, the subdifferential of $\delta_C$ at $(x_0, y_0)$ is the set of all vectors $v \in \mathbb{R}^2$ such that $$\delta_C(x,y) \geq \delta_C(x_0,y_0) + \langle v, (x-x_0, y-y_0)\rangle$$ for all $(x,y)\in C$.

Since $C$ is defined as the intersection of three sets $C_1$, $C_2$, and $C_3$, we can find the subdifferential of $\delta_C$ at $(x_0, y_0)$ by finding the subdifferentials of $\delta_{C_1}$, $\delta_{C_2}$, and $\delta_{C_3}$ at $(x_0, y_0)$ and taking their intersection.

We start by finding the subdifferential of $\delta_{C_1}$ at $(x_0, y_0)$. Since $C_1$ is a closed convex set, its subdifferential at any point is the set of all vectors pointing inwards from the boundary of $C_1$ at that point. Thus, if $(x_0, y_0)$ is a point on the boundary of $C_1$, the subdifferential of $\delta_{C_1}$ at $(x_0, y_0)$ is the set of all vectors pointing inwards from the boundary of the unit disc at $(x_0, y_0)$.

Next, we find the subdifferential of $\delta_{C_2}$ at $(x_0, y_0)$. Since $C_2$ is defined by the inequality $-x \leq 0$, its indicator function $\delta_{C_2}(x,y)$ is equal to $0$ if $x \geq 0$ and $-\infty$ if $x < 0$. Therefore, if $(x_0, y_0)$ is a point on the boundary of $C_2$, its subdifferential is the set of all vectors $(v_1,v_2)$ satisfying $$v_1 \geq 0, v_1(x-x_0) + v_2(y-y_0) \leq 0 \text{ for } x < x_0.$$

Finally, we find the subdifferential of $\delta_{C_3}$ at $(x_0, y_0)$. Since $C_3$ is defined by the inequality $-y \leq 0$, its indicator function $\delta_{C_3}(x,y)$ is equal to $0$ if $y \geq 0$ and $-\infty$ if $y < 0$. Therefore, if $(x_0, y_0)$ is a point on the boundary of $C_3$, its subdifferential is the set of all vectors $(v_1,v_2)$ satisfying $$v_2 \geq 0, v_1(x-x_0) + v_2(y-y_0) \leq 0 \text{ for } y < y_0.$$

(2) Let $F(x, y)=f(x)+g(y)$ where $f, g: \mathbb{R}^1 \rightarrow \mathbb{R}$.
(a) Prove that $\partial F\left(x_0, y_0\right)=\partial f\left(x_0\right) \times \partial g\left(y_0\right)$. Recall that the product of two sets $A \times B={(a, b) \mid a \in A, b \in B}$.
(b) Use the formula in part (a) to compute $\partial F\left(x_0, y_0\right)$ for the function $F(x, y)=|x|+y^2$.

(a) To prove that $\partial F(x_0,y_0)=\partial f(x_0)\times \partial g(y_0)$, we need to show that for any $(p,q) \in \partial f(x_0) \times \partial g(y_0)$, we have $(p,q) \in \partial F(x_0,y_0)$, and conversely, for any $(u,v) \in \partial F(x_0,y_0)$, we have $u \in \partial f(x_0)$ and $v \in \partial g(y_0)$.

First, let $(p,q) \in \partial f(x_0) \times \partial g(y_0)$. This means that there exist $x_1 \in \mathbb{R}$ and $y_1 \in \mathbb{R}$ such that $$f(x) \geq f(x_0) + p(x-x_0) \quad \text{for all } x \in \mathbb{R}$$ and $$g(y) \geq g(y_0) + q(y-y_0) \quad \text{for all } y \in \mathbb{R}.$$ Adding these two inequalities, we get $$F(x,y) = f(x) + g(y) \geq f(x_0) + p(x-x_0) + g(y_0) + q(y-y_0) = F(x_0,y_0) + p(x-x_0) + q(y-y_0)$$ for all $(x,y) \in \mathbb{R}^2$. This shows that $(p,q) \in \partial F(x_0,y_0)$.

Conversely, let $(u,v) \in \partial F(x_0,y_0)$. This means that there exists $(x_1,y_1) \in \mathbb{R}^2$ such that $$F(x,y) \geq F(x_0,y_0) + u(x-x_0) + v(y-y_0) \quad \text{for all } (x,y) \in \mathbb{R}^2.$$ Setting $y = y_0$ in this inequality and using the fact that $F(x,y) = f(x) + g(y)$, we get $$f(x) + g(y_0) \geq f(x_0) + u(x-x_0) + g(y_0) + v(y-y_0) \quad \text{for all } x \in \mathbb{R}.$$ Rearranging, we get $$f(x) \geq f(x_0) + u(x-x_0) \quad \text{for all } x \in \mathbb{R}.$$ This shows that $u \in \partial f(x_0)$. Similarly, setting $x = x_0$ in the inequality $F(x,y) \geq F(x_0,y_0) + u(x-x_0) + v(y-y_0)$ and using the fact that $F(x,y) = f(x) + g(y)$, we get $v \in \partial g(y_0)$. Therefore, we have shown that $(u,v) \in \partial f(x_0) \times \partial g(y_0)$.

Combining the two parts, we conclude that $\partial F(x_0,y_0) = \partial f(x_0) \times \partial ## 数学代写|抽象代数Abstract Algebra代考 问题 1. Let$n \geq 1$be arbitrary. Consider the group$G=\operatorname{Aut}\left(\mathbb{Z}_n\right)$. (a) Find a group we have seen in this course isomorphic to$G$. (b) Let$F: G \rightarrow \mathbb{Z}_n$be the function defined by$F(g)=g(1)$. Explain why$F$is injective. (c) Let$H \subseteq \mathbb{Z}_n$be the image of$F(G)$. Explain why$H$is not a subgroup of$\mathbb{Z}_n$. (d) Find an example where$H$is isomorphic to a subgroup of$\mathbb{Z}_n$. (d) Find an example where$\operatorname{Aut}\left(\mathbb{Z}_n\right)$is isomorphic to$\operatorname{Aut}\left(\mathbb{Z}_m\right)$but$m \neq n$. (a) The group$\mathbb{Z}_n^{\times}$is isomorphic to$G$. This is because an automorphism of$\mathbb{Z}_n$is completely determined by where it sends the generator$1$, which must be mapped to another generator, i.e., an element$a$coprime to$n$. Conversely, any such$a$gives rise to an automorphism of$\mathbb{Z}_n$by mapping$1$to$a$and extending the mapping multiplicatively. (b) Suppose$F(g_1) = F(g_2)$, where$g_1, g_2 \in G$. This means$g_1(1) = g_2(1)$, so$g_1-g_2$maps$1$to$0$, which implies that$n$divides$(g_1-g_2)(k)$for all$k\in \mathbb{Z}_n$. But$n$is coprime to$1$, so$n$must divide$g_1-g_2$. Therefore,$g_1=g_2$, and$F$is injective. (c)$H$is not a subgroup of$\mathbb{Z}_n$because it is not closed under addition. For example, if$n=4$and$G\cong \mathbb{Z}_2$, then$H={0,2}$, but$1+1=2\notin H$. (d) Let$n=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$be the prime factorization of$n$, where$p_i$are distinct primes. Let$H=\langle 1+p_1\rangle \times \langle 1+p_2^{k_1}\rangle \times \cdots \times \langle 1+p_r^{k_1}\rangle \subseteq \mathbb{Z}n$, where$\langle a \rangle$denotes the subgroup generated by$a$. Then$H$is isomorphic to$\mathbb{Z}{p_1} \times \mathbb{Z}{p_2^{k_2}} \times \cdots \times \mathbb{Z}{p_r^{k_r}}$, which is a subgroup of$\mathbb{Z}_n$, and$F(G) \cong H$. (e) Let$n=4$and$m=6$. Then$\operatorname{Aut}(\mathbb{Z}_n) \cong \mathbb{Z}_2$and$\operatorname{Aut}(\mathbb{Z}_m) \cong S_3$, the symmetric group on$3$elements. These groups are isomorphic because they both have order$2$, but$m\neq n$. 问题 2. Consider the group$G=\mathbb{Z}^2$. (a) Prove that$H={(3 m, 2 n): m, n \in \mathbb{Z}}$is a subgroup. (b) Find two representative elements from each of the cosets of$H$. (c) Plot the elements of$\mathbb{Z}^2$as points in the plane. Colour each of the cosets of$H$in a different colour. (d) Describe a group isomorphism from$G$to$H$. (e) Is this an automorphism? Explain. (a) To show that$H$is a subgroup of$G$, we need to verify the following: • The identity$(0,0)$is in$H$. • If$(3m_1,2n_1)$and$(3m_2,2n_2)$are in$H$, then their sum$(3m_1+3m_2,2n_1+2n_2)=(3(m_1+m_2),2(n_1+n_2))$is also in$H$. • If$(3m,2n)$is in$H$, then its inverse$(-3m,-2n)=(-3(m),-2(n))$is also in$H$. All of these are true, so$H$is a subgroup of$G$. (b) Let$(3m,2n)$be an arbitrary element of$G$. Then we have $$(3m,2n)=(3,2)(m,n)+(-3,-2)(m,n)+(0,0).$$ Thus,$(3m,2n)$is in the same coset as$(0,0)$,$(3,2)$, or$(-3,-2). Representative elements from each coset are: \begin{aligned} (0,0)+H &={(3m,2n):m, n \in \mathbb{Z}}, \ (3,2)+H &={(3m+1,2n+1):m, n \in \mathbb{Z}}, \ (-3,-2)+H &={(3m-1,2n-1):m, n \in \mathbb{Z}}. \end{aligned} (c) Here is a plot of\mathbb{Z}^2$, with each coset of$H$colored differently: (d) Define the function$\varphi: G \rightarrow H$by$\varphi(m,n)=(3m,2n)$. We claim that$\varphi$is an isomorphism. To see that$\varphi$is a homomorphism, observe that $$\varphi((m_1,n_1)+(m_2,n_2))=\varphi((m_1+m_2,n_1+n_2))=(3(m_1+m_2),2(n_1+n_2))=(3m_1,2n_1)+(3m_2,2n_2)=\varphi(m_1,n_1)+\varphi(m_2,n_2).$$ To see that$\varphi$is injective, suppose$\varphi(m_1,n_1)=\varphi(m_2,n_2)$. Then$3m_1=3m_2$and$2n_1=2n_2$, which implies$m_1=m_2$and$n_1=n_2$. Therefore,$(m_1,n_1)=(m_2,n_2)$, and$\varphi$is injective. To see that$\varphi$is surjective, let$(3m,2n)$be an arbitrary element of$H$. Then$\varphi(m,n)=(3m,2n)$, so$\varphi$is surjective. Therefore,$\varphi$is an isomorphism from$G$to$H$. (e) This is not an automorphism, since$G$and$H$have different. ## 数学代写|波动率模型Volatility Model代考 问题 1. 2 Download matlab Brownian motion model from the courseworks. Modify it to Geometric Brownian motion with starting value$X_o=80$, growth rate$\mu=0.05$, volatility$\sigma=0.22and 5000 trajectories. Check that the code works. Try out 50,000 trajectories. Try out 100,000 trajectories. Submit the code printout and the graph printout for 5000 trajectories. To modify the Matlab code for Brownian motion to Geometric Brownian motion, we need to modify the formula for calculating the next value of the stock price. In Brownian motion, the change in the stock price is determined by a normal distribution with mean 0 and standard deviation σ times the square root of the time step Δt. In Geometric Brownian motion, the change in the stock price is determined by a normal distribution with mean (μ – σ^2/2) times Δt and standard deviation σ times the square root of Δt. The formula for calculating the next value of the stock price becomes: $$X_{t+1} = X_t * e^{( \mu – \frac{\sigma^2}{2} ) \Delta t + \sigma \sqrt{\Delta t} * Z}$$ where Z is a standard normal random variable. 问题 2. Using arbitrage arguments explain why the price of an American call option on a stock paying no dividends should be the same as the price of a corresponding European call. Why American calls on a nondividend paying stock should not be exercised early. An American call option gives the holder the right to buy an underlying stock at a strike price K at any time up to the expiration date T of the option. A European call option gives the holder the same right, but only at the expiration date T. If we assume that the stock pays no dividends, then the stock price S follows a geometric Brownian motion with constant drift μ and volatility σ, and the risk-free interest rate is denoted by r. Let C and C_A denote the prices of a European call and an American call on the stock, respectively. We want to show that C = C_A. Suppose for contradiction that C > C_A. Then, an arbitrage opportunity arises. Here’s how to exploit it: 1. Buy the European call option for price C. 2. Simultaneously, sell (short) the stock for price S and invest the proceeds at the risk-free rate r. This creates a cash inflow of S – C. 3. At expiration T, the value of the European call option is max(S(T) – K, 0), which we can use to buy the stock if it’s in the money (S(T) > K), or keep the cash if it’s out of the money (S(T) ≤ K). 4. If the European call option is exercised, we use the stock we bought at step 2 to fulfill our obligation and deliver it at the strike price K, receiving K in return. If the European call option is not exercised, we keep the stock we bought at step 2 and sell it for S(T) in the market, receiving S(T) in return. 5. In either case, we end up with a profit of S(T) – C – (S – C)e^(rT), which is greater than or equal to zero by construction. Therefore, the assumption that C > C_A leads to an arbitrage opportunity, which cannot exist in an efficient market. Therefore, we must have C = C_A. Now, why should an American call option on a nondividend paying stock not be exercised early? Suppose that the option is in the money (S > K) and the holder exercises the option early. This means that the holder pays the strike price K and receives the stock, which is worth more than K. But this transaction can also be achieved by selling the option in the market, which is worth at least as much as the stock, and then buying the stock at the market price, which is less than or equal to S. Therefore, exercising the option early leads to a strictly lower payoff than selling the option in the market and buying the stock at the market price. Therefore, it’s never optimal to exercise an American call option on a nondividend paying stock early. ## 数学竞赛代考|PROMYS 2023 Application Problem Set代写 问题 1. 1. Consider the sequence \begin{aligned} & a_1=2^1-3=-1, \ & a_2=2^2-3=1 \ & a_3=2^3-3=5 \ & a_4=2^4-3=13 \ & \vdots \ & a_n=2^n-3 \ & \vdots \end{aligned} defined for positive integersn$. Which elements of this sequence are divisible by 5 ? What about 13 ? Are any elements of this sequence divisible by$65=5 \cdot 13$? Why or why not? To check if an element of the sequence is divisible by 5 or 13, we simply need to check if$a_n$is congruent to 0 modulo 5 or 13, respectively. First, we will find the remainder of each$a_nwhen divided by 5: \begin{aligned} a_1 &\equiv -1 \equiv 4 \pmod{5}, \\ a_2 &\equiv 1 \pmod{5}, \\ a_3 &\equiv 5 \equiv 0 \pmod{5}, \\ a_4 &\equiv 13 \equiv 3 \pmod{5}, \\ a_5 &\equiv 29 \equiv 4 \pmod{5}, \\ a_6 &\equiv 61 \equiv 1 \pmod{5}, \\ a_7 &\equiv 125 \equiv 0 \pmod{5}, \\ a_8 &\equiv 253 \equiv 3 \pmod{5}, \\ a_9 &\equiv 509 \equiv 4 \pmod{5}, \\ a_{10} &\equiv 1021 \equiv 1 \pmod{5}, \\ &\vdots \\ a_{13} &\equiv 8193 \equiv 3 \pmod{5}, \\ a_{14} &\equiv 16381 \equiv 1 \pmod{5}, \\ a_{15} &\equiv 32765 \equiv 0 \pmod{5}, \\ &\vdots \end{aligned}a1​a2​a3​a4​a5​a6​a7​a8​a9​a10​a13​a14​a15​​≡−1≡4(mod5),≡1(mod5),≡5≡0(mod5),≡13≡3(mod5),≡29≡4(mod5),≡61≡1(mod5),≡125≡0(mod5),≡253≡3(mod5),≡509≡4(mod5),≡1021≡1(mod5),⋮≡8193≡3(mod5),≡16381≡1(mod5),≡32765≡0(mod5),⋮​ From the above calculations, we see thata_3$,$a_7$,$a_{12}$,$a_{15}$,$a_{18}$,$a_{23}$,$a_{27}$,$a_{30}$, and so on, are divisible by 5. Next, we will find the remainder of each$a_nwhen divided by 13: \begin{aligned} a_1 &\equiv -1 \equiv 12 \pmod{13}, \\ a_2 &\equiv 1 \pmod{13}, \\ a_3 &\equiv 5 \pmod{13}, \\ a_4 &\equiv 13 \equiv 0 \pmod{13}, \\ a_5 &\equiv 29 \equiv 3 \pmod{13}, \\ a_6 &\equiv 61 \equiv 9 \pmod{13}, \\ a_7 &\equiv 125 \equiv 12 \pmod{13}, \\ a_8 &\equiv 253 \equiv 1 \pmod{13}, \\ a_9 &\equiv 509 \equiv 5 \pmod{13}, \\ a_{10} &\equiv 1021 \equiv 4 \pmod{13}, \\ &\vdots \\ a_{12} &\equiv 4093 \equiv 0 \pmod{13}, \\ a_{13} &\equiv 8193 \equiv 1 \pmod{13}, \\ a_{14} &\equiv 16381 \equiv 3 \pmod{13}, \\ a_{15} &\equiv 32765 \equiv 5 \pmod{13}, \\ &\vdots \end{aligned}a1​a2​a3​a4​a5​a6​a7​a8​a9​a10​a12​a13​a14​a15​​≡−1≡12(mod13),≡1(mod13),≡5(mod13),≡13≡0(mod13),≡29≡3(mod13),≡61≡9(mod13),≡125≡12(mod13),≡253≡1(mod13),≡509≡5(mod13),≡1021≡4(mod13),⋮≡4093≡0(mod13),≡8193≡1(mod13),≡16381≡3(mod13),≡32765≡5(mod13),⋮​ From the above calculations, we see thata_4$,$a_7$,$a_{12}$,$a_{13}$,$a_{14}$,$a_{15}$,$a_{16}$,$a_{19}$,$

1. To get the echo of a positive integer, we write it twice in a row without a space. For example, the echo of 2023 is 20232023 . Is there a positive integer whose echo is a perfect square? If so, how many such positive integers can you find? If not, explain why not.

Let $n$ be a positive integer and let $m$ be the number of digits in $n$. Then the echo of $n$ is given by

n\cdot 10^m+n.n⋅10m+n.

We want to find positive integers $n$ such that the above expression is a perfect square.

Let $k$ be a positive integer such that

n\cdot 10^m+n=k^2.n⋅10m+n=k2.

Rearranging, we have

n(10^m+1)=k^2.n(10m+1)=k2.

Since $10^m+1$ is odd and $n$ and $k^2$ have the same parity, we conclude that $n$ is odd.

Now, let us consider the prime factorization of $10^m+1$. If $m$ is even, then $10^m+1$ is the sum of two squares and therefore cannot have a prime factor of the form $4k+3$. Hence, all prime factors of $10^m+1$ are congruent to 1 modulo 4. If $m$ is odd, then $10^m+1$ is divisible by 11. In either case, the prime factors of $10^m+1$ are finite and all congruent to 1 modulo 4.

Therefore, if $n$ is odd and the expression $n(10^m+1)$ is a perfect square, then the prime factorization of $n$ must consist of primes that are congruent to 1 modulo 4, and the exponents of these primes must be even. Moreover, the prime factorization of $10^m+1$ also consists of primes that are congruent to 1 modulo 4, and the exponents of these primes must be even.

Let $n=p_1^{2a_1}\cdots p_k^{2a_k}$ be the prime factorization of $n$ as above, and let $10^m+1=q_1^{2b_1}\cdots q_l^{2b_l}$ be the prime factorization of $10^m+1$ as above. Then we have

n(10^m+1)=p_1^{2a_1}\cdots p_k^{2a_k}q_1^{2b_1}\cdots q_l^{2b_l}.n(10m+1)=p12a1​​⋯pk2ak​​q12b1​​⋯ql2bl​​.

Since $n$ and $10^m+1$ are relatively prime, it follows that the primes $p_1,\ldots,p_k$ and $q_1,\ldots,q_l$ are distinct. Moreover, $p_1,\ldots,p_k$ are congruent to 1 modulo 4, as are $q_1,\ldots,q_l$.

Since $n$ and $10^m+1$ have the same parity, it follows that $p_1,\ldots,p_k,q_1,\ldots,q_l$ are all odd.

Now, let us consider the exponent of 2 in the prime factorization of $n(10^m+1)$. Since $n$ is odd, the exponent of 2 in the prime factorization of $n$ is 0. Since $10^m+1$ is odd, the exponent of 2 in the prime factorization of $10^m+1$ is also 0. Therefore, the exponent of 2 in the prime factorization of $n(10^m+1)$ is 0.

## Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

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