# Tag Info

Consider a discrete random variable $X$ taking on values $x_1, x_2, \cdots, x_n$ with positive probabilities $p_1, p_2, \ldots, p_n$ respectively. Two shibboleths that statisticians don't just murmur but instead shout from the rooftops are that probabilities are nothing but long-term frequencies, and that an event of probability $p$ will occur approximately $... 7 It's analogous to the discrete version. It's generally useful to think$P(X=x)\approx f(x)\Delta x$in the continuous case. In the limiting case, as$\Delta x$goes to$0$, this probability is$0$. So, the expected value will be $$E[X]\ \approx \sum x\ P(X=x)\ =\ \sum_{x\in\{a,a+\Delta x,...,b\}} x\ f(x)\Delta x$$ If you take the limiting case, this is ... 1 Hello and welcome to this community. In order to clarify your doubts, you can first think of discrete spaces. Take for example the random experiment concerning the toss of a fair dice. In this case, the sample space (that just mean the set of all possible outcomes) is$\Omega = \{1,2,3,4,5,6\}$, and we have a probability measure$\mathbb{P}$that assigns to ... 0 Maybe I'm misunderstanding the question but I fail to see why that expectation doesn't just blow up to infinity. Intuitively it is essentially just counting the number of times you draw red, given you have an infinite number of draws. And the only way this kind of infinite sum converges is if it is a geometric sequence inside some radius of convergence. But ... 0 If you're reading DSML, then I think you can just use the result they proved in the text for squared-error loss: note that$\mathbb{1}[Y_i \neq \hat{Y}_i] = (Y_i - \hat{Y}_i)^2$when$Y_i, \hat{Y}_i \in \{0,1\}$. 5 As here, truncating a standard normal distribution between$a$and$bgives a random variable with density $$\dfrac{\exp\{-x^2/2\}}{\sqrt{2\pi}[\Phi(b)-\Phi(a)]}.$$ The expectation you are looking for therefore is \begin{align*} &\dfrac{1}{\sqrt{2\pi}[\Phi(b)-\Phi(a)]}\int_a^b x^3\exp\{-x^2/2\} \,dx \\ =& \dfrac{1}{\sqrt{2\pi}[\Phi(b)-\Phi(a)]}\... 2 Assuming \tau>0, let \xi=x-\mu and \lambda=\nu-\mu. Then dx=d\xi, and we have \begin{align*} \int_{-\infty}^{\infty}(x-\mu)^2 e^{-|x-\nu|/\tau}\,dx &=\int_{-\infty}^{\infty}\xi^2 e^{-|\xi+\mu-\nu|/\tau}\,d\xi\\ &=\int_{-\infty}^{\infty}\xi^2 e^{-|\xi-\lambda|/\tau}\,d\xi\\ &=\int_{-\infty}^\lambda \xi^2 e^{(\xi-\lambda)/\tau}\,d\xi +\... 1 If you have data X_1, X_2, \dots, X_n randomly sampled from \mathsf{Norm}(\mu, \sigma), with \mu known and \sigma unknown, then V = \frac{1}{n}\sum_{i=1}^n (X_i - \mu)^2 has E(V) = \sigma^2, with \frac{nV}{\sigma} \sim \mathsf{Chisq}(\nu = n). The proof follows directly from the definition of \mathsf{Chisq}(\nu = n) as the distribution of the ... 1 A Bayesian point estimator is a summary of the posterior used for a specific purposed and justified by an optimisation principle associated with a utility function U. The point estimate \hat\theta(\cdot) is the function that maximises the posterior expected utility\mathbb E[U(\hat\theta,\theta)|x]=\int U(\hat\theta,\theta)\,\pi(\theta|x)\,\text d\... 3 IfX_1,\ldots,X_n$are i.i.d$\mathsf{Bernoulli}(p)$and$S_n=\sum\limits_{i=1}^n X_i$, the conditional expectation$E\left[S_lS_m\mid S_n\right]$is just the unbiased estimator of$E\left[S_lS_m\right]$based on$S_n$by Lehmann-Scheffé theorem. In other words, it is the UMVUE of$E\left[S_lS_m\right]$. Keeping in mind that$l<m, \begin{align} E\left[... 2 No it wouldnt apply. The statement E[XY]=E[X]E[Y] is only valid when X and Y are independent, which is not the case in your setting since the summations run over the same X_i variable, making the sums dependent. The answer here may be useful: https://math.stackexchange.com/questions/1476906/expectation-of-the-product-of-two-dependent-binomial-random-variable 2 This is known as model-averaged quantile regression. It was discussed in Koenker "A note on L-estimates for linear models" (1984), though its origins stem from as early as the seminal paper of Koenker & Bassett "Regression quantiles" (1978). The 1984 paper actually proposes composite quantile regression, a natural competitor/... 1 I agree that this analysis (for the normal distribution with known variance) is not particularly well explained in the book. Looking at why\mathrm{E}(\mathrm{E}(\tilde{y} | \theta, y) | y)=\mathrm{E}(\theta | y)$, first consider$\mathrm{E}(\tilde{y} | \theta, y)$. By definition of the assumed model, two observations$y$and$\tilde{y}$are independent ... 0 Normal distribution and shifted Poisson are examples. The shifted Poisson is$s=x-\lambda$, where$\lambda$is Poisson intensity. There's a whole family of distribution such that the linear combination (not just the sample mean) of variables follows the same distribution, it's called stable distribution. 19 You can think about your problem as a Markov chain, i.e., a set of states with certain transition probabilities between states. You start in one state (all cards face up) and end up in an absorbing state (all cards face down). Your question is about the expected number of steps until you reach that absorbing state, either for a single chain, or for the ... 11 I think I've found the answer for the single player case: If we write$e_{i}$for the expected remaining length of the game if$i$cards are facedown, then we can work out that: (i).$e_{5} = \frac{1}{6}(1) + \frac{5}{6}(e_{4} + 1)$(ii).$e_{4} = \frac{2}{6}(e_{5} + 1) + \frac{4}{6}(e_{3} + 1)$(iii).$e_{3} = \frac{3}{6}(e_{4} + 1) + \frac{3}{6}(e_{2} + 1)$... 1 First of all,$P(0<X<b)=b$not$x/b$, i.e. not a function of$x$. Similarly,$P(a<X<1)=1-a$. These are also equal to$E[Y],E[Z]$respectively. Also, we have$E[Y^2]=E[Y],E[Z^2]=E[Z]$for binary RV case. For the joint moment,$E[YZ]=P(Y=1\cap Z=1)=P(a<X<b)=b-a$. I think you can follow from here. 1 tl;dnr version: The OP's final answer$E[Z] = 2-p$is correct but the reasoning is not. The book's statement that$Z$does not have a normal density is correct, but its mean computation incorrect (perhaps a typo). As Chris Haug says, the OP's first statement (which the OP has deleted a few minutes ago) is incorrect. It is not true that$Z = pX + (1-p)Y$... 0 In finance, you often use the log normal distribution to price securities, including options, which give you a right (but not an obligation) to buy a stock at a specified price by a certain date (the maturity). Option prices are functions of the price of the stock, the volatility of the stock, the so-called risk-free interest rate, and time until the ... 3 I don't think's there's a deep meaning behind differentiating a constant. I'm sure if you think long enough something can be found. However, I'd sort this into "integration tricks" bucket. It's routinely used in theoretical physics. 0 The very first line of your derivation is not correct (that's not the correct expression for the mixture$Z$). To illustrate, if$X$and$Y$are independent and if$p=0.7$, this is what these densities look like: Note that the mixture$Z$is bimodal, but$pX + (1-p)Y$is actually a linear combination of independent normal variables, and is also normal. ... 1 The computations look correct. There's no such "symmetry" in the crosscovariance function because they don't measure the same effect at all:$h=1$is the effect of today's$Y$on tomorrow's$Xh=-1$is the effect of today's$X$on tomorrow's$Y$Here,$X_t$is a predictor of$Y_{t+1}$, but$Y_t$contains no information about$X_{t+1}$at all. ... 3 Let$E_n$be the expected number of trials if the starting number is$n$. Then, $$E_n=1+\frac{1}{n}\sum_{i=1}^{n-1} E_i$$ Via some algebraical manipulations, you'll have $$E_{n}=E_{n-1}+\frac{1}{n}\rightarrow E_n=\sum_{i=1}^n\frac{1}{i}$$ So, it's not$\log_2n$. 1 The formula for the expected value of a gamma random variable (with shape parameter$\alpha$and scale parameter$\beta$) constrained to an interval$\left[ a,b \right]$can be expressed as $$E \left[ X \ | \ a<X<b \ \right]=\frac{\alpha \beta \left[ P \left( \alpha+1,\frac{b}{\beta} \right) - P \left( \alpha+1,\frac{a}{\beta} \right) \right] }{P \... 2 What you are looking for is the expectation of a truncated gamma distribution. Formulas (11) and (13) in "A Right and Left Truncated Gamma Distribution with Application to the Stars" by Zaninetti (there is a pdf here) give you the formula you are looking for. Let b denote the scale and c the shape. Then$$ E(X|X\in[x_\ell,x_u]) = b^2k\bigg(\Gamma\Big(1+... 3 Causality is by definition a special case of stationarity. Stationarity, or causality, does not imply mean 0. Where you went wrong is you're comparing different AR models---one without intercept and one with. Stationary AR models without intercept have zero mean in general, whether causal or non-causal. Vice versa for those with intercept. When you ... 3 What you are looking for is the expectation of a truncated Weibull distribution. "Truncated Weibull Distribution Functions and Moments" by François Crénin gives you the formula you need. Let$\alpha$denote the shape and$\beta$the scale of the Weibull, then $$E(X|a<X<b) = \frac{\beta}{e^{-\left(\frac{a}{\beta}\right)^\alpha}-e^{-\left(\frac{b}{\... 1 The proof you give is going to depend on how deep you need to go, but if you want to go deep into measure theory then this is essentially an application of the Carathéodory extension theorem. Specifically, you will need to apply the following theorem, which can be proved using the Carathéodory extension theorem. (I leave the proof as an exercise for you.) ... 0 Same as we can use f_X(x) for the probability density f of random variable X evaluated at x, you can use notation like f_{Y|X}(y|x) or \mu_{Y|X}(y|x) to denote conditional things (e.g. densities, expected values etc.). See examples here or here. 0 As a precursor: It is worth thinking about how these problems arise in statistical practice. Optimising over x is rare - usually, x has already been observed. It is more common to be optimising over \boldsymbol{\theta}, given an observation x, e.g. to find the maximum likelihood estimator of \theta, one would solve$$\max_\boldsymbol{\theta} \left\... 2 This is how Self-Normalized Importance Sampling (SNIS) works - you draw samples from a proposal distribution that is essentially guess about where This shows how the lack of knowledge about$\log Z$can be solved. But it doesn't mean that lack of knowledge of$\log Z$is not a problem. In fact the SNIS method shows that not knowing$\log Z$is a problem. It ... 2 Let$Y$be the RV representing the time. Given$X$, we know that$Y$is uniform with mean$E[Y|X]=\frac{3}{2}\sqrt{X}$. Using the law of total expectation, $$E[Y]=E[E[Y|X]]=\frac{3}{2}E[\sqrt{X}]$$ which you can find easily with the marginal distribution of$X\$. Edit I don't recommend this way, but it's good practice for you to verify. Here is the marginal ...