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From your description, the height (denoted by $X$) of British citizens follows normal distribution with mean $\mu = 160\,cm$ and STD $\sigma = 7.6\,cm$. Denote the PDF of $X$ as $f(x)$. What you should do is to find out the value of $z$, such that $\int_{z}^{175.3} {f(x)}\,\text{d}x = 0.95$.

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Since this is self-study question, let me give you such a hint: What about trying to simulate such structure in statistical software? We can assume values and functional forms, and then check if regression gives right values. Lets try to create such structure, in this example only for variables Z3, X, W3, and Y (we cut out the rest): An R code which creates ...

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Yes, it is. As you mentioned, the classical rule is $P(A,B) = P(A|B)P(B)$, but it can also be applied to conditional probabilities like $P(\cdot|C)$ instead of $P(\cdot)$. It then becomes $$P(A,B|C) = P(A|B,C)P(B|C)$$ (you just add a condition on $C$, but otherwise that's the same formula). You can then apply this formula for $A = y$, $B = \theta$, and $C ... 1 Your solution seems to be correct. The strange shape of your parameter space (it's not a open subset of$\mathbb R^2$) creates this ambiguity in the final result: each combination of$(p,q,r)$gives a different LRT. Some are more powered for$\mu_1=0$, some are more powered for$\mu_2=0$and some are more powered for$\mu_1\neq0,\mu_2\neq0$, but all of them ... 0 Suppose that$X_{1},\ldots,X_{n}$are iid exponential random variables, with density function$f(x;\theta)=\theta e^{-\theta x}$. Then the likelihood function will be \begin{equation*} \text{L}(\theta|x)=\prod_{i=1}^{n}f(x_{i};\theta)=\prod_{i=1}^{n}\theta e^{-\theta x_{i}}=\theta^{n} e^{-\theta n\bar{x}} \end{equation*} where$n\bar{x}=\sum_{i=1}^{n}x_{i}.$... 0 You seem confused about the concepts of sufficiency and unbiasedness. To be clear: unbiasedness plays no part in the definition of sufficiency. The reason is simple: If$T$is sufficient for some parameter$\theta$indexing a distribution family, then so is also$T/10, 10T, 1000T$and a lot of other one-to-one functions of$T$. Al those functions cannot be ... 1 In accordance with whuber's suggestion, I am posting an extended version of some comments that I made on whuber's answer as a separate answer of my own. The experiment consists of players A and B each (independently) tossing their individual coins that turn up Heads with probabilities$p_A$and$p_B$respectively. Repeated independent trials of this ... 2 This is actually a bayesian problem. Time$Y_1$if you don't know the number of mails at time 1, is exponentially distributed, you got that right. But when you get the additional information that at time 1 Alice only sent one e-mail, you have to update your distribution of$Y_1$. Applying Bayes rule: $$p(Y_1|emails_1 = 1) \propto p(emails_1 = 1|Y_1)p(Y_1).$$ ... 1 I think you may be confusing$\hat \beta_i$with$\beta_i$and$b_i$:$\beta_i$- unknown parameter$\hat \beta_i$and$b_i$is the same thing - an estimator for an unknown parameter. Hint 1: Try to use only$\beta_i$and$\hat \beta_i$Hint 2: Then take look at the test statistic. 2$X \sim Bern(0.2)$By the definition of median$P(X \leq m) \geq 1/2$and$P(X \geq m) \geq 1/2$It has$m = \begin{cases} 0, \quad p < 1/2\\ [0,1], \quad p = 1/2\\ 1, \quad p > 1/2 \end{cases}$It then follows that$m = 0$. 6 Let$X \sim \mathsf{Bern}(p=.2)\equiv\mathsf{Binom}(n=1, p=.2).$In R, where qbinom is the inverse CDF (quantile function) of a binomial distribution a median$\eta = 0.$qbinom(.5, 1, .2)  0$P(X \le 0) = P(X = 0) = 0.8 \ge 1/2.$dbinom(0, 1, .2)  0.8 And obviously,$P(X \ge 0) = 1 \ge 1/2.$The CDF of$X$is plotted below. The median of$X$is ... 2 For the first question, since$Y_{n}$is of binomial distribution, the probability that$Y_{n}=y$is$P(Y_{n}=y)={n \choose y}\theta ^y\times(1-\theta)^{n-y}$. The probability that$n-Y_{n}=k$then is$P(n-Y_{n}=k)=P(Y_{n}=n-k)={n \choose n-k}\theta ^{n-k}\times(1-\theta)^{k}={n \choose k}(1-\theta) ^{k}\times\theta^{n-k}$. Thus,$n-Y_{n}$is of binomial ... 2 For you second question, since you already know that the mean of$Y_{n}$is$n\theta$, and$n\sim\mathcal{Poisson}(\lambda)$, the mean of$Y_{n}$is$E[Y_{n}] = E[n\theta]=\theta E[n]=\lambda\theta$. Or equivalently,$E[Y_{n}]=\sum_{n} n\theta*\frac{\lambda^{n} e^{-\lambda}}{n!}=\lambda\theta$. Similarly, the various of$Y_{n}$is$\sum_{n} n\theta (1-\theta)...

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The 2x2 transition matrix consists of 4 probabilities: $p(s|s)$, $p(d|s)$, $p(s|d)$ and $p(d|d)$, where $s$ and $d$ stand for "same votes" and "different votes" respectively. In case when Alice and Bob had different (previous) votes, then if both switch or both do not switch their choices, they will have different votes again. Otherwise, ...

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Here is one unnecessary way to approximate the answer to this question, using normal distributions. The expected number of cars $X$ is $E[X]=0.25 \cdot 0+0.5 \cdot 1+0.25 \cdot 2=1$. The standard deviation of $X$ is $SD[X]=\sqrt{(0-1)^2 \cdot 0.25+(1-1)^2 \cdot 0.5+(2-1)^2 \cdot 0.25}=\sqrt{0.5}=0.707$. Using convolutions we can sum 100 such normal ...

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You can define auto-correlation functions for non-stationary time series. An auto-correlation like this $$\sum_{t=0}^n X_t \overline{X_{t+\tau}}$$ can always be computed for a sample. Also a correlation function in terms of the expectation can always be expressed (except if the covariance or 2nd moment is not defined which may occur with distributions that ...

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Roughly, assuming a constant population and its random mixing between the time of tagging and the time of observing hawks at the feeder, the proportion $10/N$ of tagged hawks in the population should be estimated by the proportion $6/28.8$ at the feeder. So we estimate $\hat N = 288/6 = 48$ hawks in the population. Note: This 'Lincoln-Peterson' method fails ...

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You computed the CDF by using the proper integral of the PDF $$\int 2x^{-2} dx = \frac{-2}{x} + C$$ But what you forgot is to use the correct integration constant (or use a definite integral). Your CDF is not $$F(x) = \frac{-2}{x}$$ But instead F(x) = \begin{cases} 0 &\quad \text{if} \quad x \leq 2 \\ \int_2^x 2u^{-2} du = 2 - \frac{2}{x} &\quad ... 1 For the Gaussian model (with variance \sigma_{\epsilon}^2 = \hat{\sigma}_{\epsilon}^2 assumed known), the AIC statistic is equivalent to C_p, and so we refer to them collectively as AIC. Suppose a Gaussian model with variance \sigma_{\epsilon}^2 = \hat{\sigma}_{\epsilon}^2 = \sigma^2 Mallow's C_p [ C_p = \frac{1}{n} (RSS + 2 d \sigma^2) ] AIC is \... 3 Hints: As you know and state clearly, Leibniz's rule says something like If F(\theta) = \displaystyle\int_{a(\theta)}^{b(\theta)} f(x; \theta) \,\mathrm dx where a(\theta), b(\theta), and f(x; \theta) are differentiable functions of \theta, then \begin{align}\frac{\mathrm dF(\theta)}{\mathrm d\theta} &= \frac{\mathrm d}{\mathrm d\theta}\int_{a(\... 3 It follows from the law of iterated expectations: the expected value of the conditional expected value of u given X is the same as the expected value of u.E[u^2] = E[E[u^2|X]] = E[\sigma^2] = \sigma^2$$4 Bayesian inference can be used in any scenario where you could use other forms of statistical inference, e.g. maximum likelihood. Additionally, it has some extra advantages, since it allows you for using priors, so for bringing out-of-data information into the model, and it gives you uncertainty estimates for free, since you learn the distribution of the ... -1 I guess you are confused because it did not mention an exact number. If the question said something like rolling a die twice and getting a 1 twice then you are correct but since it did not say so, the same is true for getting a 1,2 3 or any other number. Thus since there are 6 numbers that can turn up that way, it will be 6/36 which will give you 1/6. 2 By remembering how the geometric distribution arises, we can solve this problem with almost no calculation. The problem can be seen as a competition A geometric random variable W models the number of failures in a sequence of independent Bernoulli trials before the first success is observed. Its parameter p is the chance of success in each trial. The ... -2 As I indicated in a comment above there is a cited path to the solution available here courtesy of the Math forum on Stack Exchange on the topic 'Difference between two independent geometric distribution. However, I have also found another route that is worthy of mention as it may have more broad applications (that is, for other distributions than the ... 0 Hint:$$ \begin{align} P(X<-b) + P(-b < X \leq b) + P(X \geq b) & = 1 \\ P(-b < X \leq b) & = 1 - \big(P(X<-b) + P(X \geq b)\big) \end{align} $$Since normal distribution is symmetric around X=0, the right-hand side of the equation can be simplified into$$ \begin{align} P(-b < X \leq b) & = 1 - \big(P(X < -b) + P(X < -b)\...

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Even simpler. There are eight possible outcomes (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT)and only one does not contain a heads. If you stop after one or two throws makes no difference.

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Unless I am missing something, this is a very basic problem. You are throwing 1 coin 3 times and looking for the probability they will all come up heads, correct? This is dictated by the binomial distribution. But simply, p(heads) on toss 1 = 0.5. Probability that toss 1 and toss 2 come up heads = 0.5*0.5 = 0.250. Probability that toss 1, 2, and 3 all ...

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Rather than calculating it as stopping flipping a coin once that coin comes up heads, you should calculate the probability based on flipping each coin three times first, and then looking at whether any of the results were heads.

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Ah, I see you are studying Pearl's Causal Inference in Statistics: A Primer, co-written with Glymour and Jewell. Excellent choice! Theorem 4.3.2 says that if $\tau$ is the total effect of $X$ on $Y$, which is $ab$ in this case, then, for any evidence $Z=e,$ we have $$E[Y_{X=x}|Z=e]=E[Y|Z=e]+\tau(x-E[X|Z=e]).$$ So, in our case, if we want to compute $E[Y_1-... 1 To start, look at the rule for$Var(aX + b)$in terms of$Var(X).$If$Q\sim\mathsf{Chisq}(\nu=1),$then$Var(Q) = 2$and$Var(Q/2)=?,\;Var(Q/\sqrt{2}) = ?$But neither$Q/2$nor$Q/\sqrt{2}$has a chi-squared distribution. 0 So there are 512 ways to get results when you toss 3 coins, that looks good. I think you're just missing a few nuances. It would help to the totals for each of the initial results (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT) and make sure each one totals to 1/8th. For example you have to results for an initial throw of HHT: HHT, H = (1/16) HHT, T, H = (1/32) ... 6 You can split the problem into two (independent) parts so it becomes easier to solve. Probability of flipping head for a coin after three attemps, the inverse of the probability of getting 3 tails in a row. Probability of flipping all three coins after three attemps, it is the same as the probability of flipping head for a coin after three attemps (i.e. ... 1 You can think about it as trying to flip heads with one coin with three attempts. After one attempt, the chance for H is 1/2. After two attempts (that is, you get T, and then H), the chance is 1/4. After three attempts (T, T, H), the chance is 1/8. Add it all up and the chance that you win this minigame is 7/8. In reverse, you lose by flipping T, T, T, ... 23 There is a slightly easier approach. Since you asked not to be given the answer, here are some hints: In effect you flip each coin up to three times. If it comes up heads on any of those then you stop with that coin What is the probability you get three tails with a particular coin? So what is the probability you get that coin showing heads in the up-to-... 11 The answer by Henry +1 (and others as well) is much simpler. However, the method below is more general. If the problem is changed slightly, for instance that the probability of a coin landing heads becomes dependent on the number of coins being flipped, then the simple answer does not work anymore. I would model this with a Markov chain. Giving a matrix to ... 0 You'll take the limit of the expression and apply L'hopital's rule: $$\lim_{t\rightarrow0}\frac{\log M_X(t)}{t}=\lim_{t\rightarrow0} \frac{M_X'(t)}{M_X(t)}=\frac{M'_X(0)}{M_X(0)}=\frac{\mathbb E[X]}{1}=mx$$ P.S. Please use$\LaTeX$as suggested in the comments. 0 Ok, seems I completely deserve -1 here, because, shame on me, there is absolutely correct solution -- given non-convergence in D we have non-convergence in P.Namely, given non-conv. in D impleis non-conv. in P. Everything is quite straightforward. 3 Okay, I think I've figured out the answer. Funny how typing everything out tends to clarify some incorrect assumptions at the same time. I believe that my second methodology is actually correct, but that the individual probabilities that I assigned to each case were incorrect. For example, I had assigned the probability of flipping one "heads" ... 0 The above answer was correct that you ignored the numerator$p^r$, but I found the illustration is a little confusing, so here we go for a standard way to solve the problem$\begin{aligned} \lim_{r\to \infty} M_{NB}(t) & = \lim_{r\to \infty} (\frac{p}{1-(1-p)e^t})^r \\ &= \lim_{r\to \infty} (\frac{1-(1-p)}{1-(1-p)e^t})^r \\ &= \lim_{r\to \infty} \...

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This is impossible. An immediate way to see this is to write $\widehat \beta = \beta + O_p(n^{-1/2})$ whereas convergence in probability is equivalent to $\widehat \beta = \beta + o_p(1)$. But I'll outline a more rigorous argument since it seems like you may not be comfortable with this type of argument (it sort of begs the question if you don't know why $\... 1 Hint: $$P(X \ge 2\lambda) = P(X - \lambda \ge \lambda) \le P(|X-\lambda| \ge \lambda)$$ (To justify the inequality, note that$X-\lambda \ge \lambda$implies$|X-\lambda| \ge \lambda$.) Then apply Chebychev's inequality to the right-hand side. 0 Hint: there is a lot of courses branded as "statistics" or "data science", while not many as "econometrics", so try the two former keywords. The topics covered by such courses would overlap to great extent with econometrics. Moreover, such courses would cover a general set of skills for working with data, data analytics, and ... 0 Hi Phil as a fellow economics disciple, I would like to say that book is always the best material for the learning, but the online course is the complimentary or an enhancement for learning process. I give you both the Youtube channels that pretty much sum up all the econometric course for undergrad and master and the books. The links are below, if you ... 2 Two-sample test. Suppose you have two independent normal samples, each of size 11: set.seed(13) x1 = rnorm(11, 142, 21) x2 = rnorm(11, 123, 23) To test whether these two samples come from populations with the same center, you might do a Wilcoxon rank sum test. With P-value$0.01$you reject$H_0,$correctly finding that the data were sampled for ... 1 You need to get exactly 2 Fs among the first five firms audited; then the last one must be the third F. So it's${5\choose 2}(.1)^3(.9)^3 =0.00729$or, for number$X$of Fs in$n=5$audits, its$(0.1)P(X=2, 5, .1) = 0.00729.\$ [In R.] choose(5,2)*.1^3*.9^3  0.00729 dbinom(2, 5, .1)*.1  0.00729

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