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\begin{align}E[X\log X]&=\int_0^\infty \log x \frac{\lambda^\alpha}{\Gamma(\alpha)}x^\alpha e^{-\lambda x}dx\\&=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\lambda}\int_0^\infty\log x\frac{\lambda^{\alpha+1}}{\Gamma(\alpha+1)}x^{(\alpha+1)-1}e^{-\lambda x}dx\\&=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\lambda}E[\log Y]\end{align} where $Y\sim \text{... 6 You'll use law of iterated expectations: $$E[X]=E[E[X|Y]]=E[Y]=\frac{10}{10-2}=5/4$$ 3$E(Y)$is a probability theory result. That means$Y$(written in caps) is a random variable having a density function$f_Y$.$E(Y) = \int y f_Y$. The integration is the Riemann-Stieltjes integral. In other words, if$Y$is discrete, it boils down to a summation, if$Y$is continuously valued, it's a standard integral. Also in probability theory, lower case ... 2 Using the definition of variance, you have: $$\operatorname{var}(Y)=E[X^4]-E[X^2]^2$$ You need the 4-th non-central moment of normal RV. Since, the mean is$0$, you can also use central moment, which is$3\sigma^4$. Subtracting$E[X^2]^2=\sigma^4$makes variance of$Y2\sigma^4$. Another way is to use the relation between normal, chi-squared and gamma ... 1 Although$\hat\beta$is a function of both$X$and$y$,$\mathbb E[\hat\beta]$is a constant because expectation is taken over all RVs inside. Let's call it$c, and take the inside of the outermost expected value only. The expression becomes the following: \begin{align}\mathbb E[\overbrace{(\mathbb E[\hat\beta|X]-c)}^{\text{given X this is const, call it ... 1 Note that this would be a beta distribution such as the one in here with \alpha = \theta and \beta = 1. First, consider that E[x] = \int_0^1x\theta x^{\theta-1} dx= \theta \int_0^1x^\theta dx=\frac{\theta}{\theta+1} x^{\theta+1}|_0^1 = \frac{\theta}{\theta+1}. Then you can find \hat{\theta} using the method of moments such that \frac{\hat{\theta}}{... 1 Via law of iterated expectations, it's equal to\mathbb E[\max_{0<t<1} X(t)]=\mathbb E[\mathbb E[\max_{0<t<1}X(t)|A,B]]=\mathbb E[\max (A,B)]$$, which you can use the formula found in Equation (11) of this paper. Note that the answer involves the CDF function \Phi(t) of standard normal RV. 1 Although there is no closed form, if K is big enough we can use Central Limit Theorem to approximate X\sim\mathcal{N}(Kq,Kq(1-q)) and Y\sim\mathcal{N}(Kp,Kp(1-p)). It follows that pX-qY has asymptotically normal distribution with mean zero and variance given by:$$\sigma^2=p^2\cdot Kq(1-q)+q^2\cdot Kp(1-p) \\=Kpq\left[p(1-q)+q(1-p)\right]\\=Kpq(p+q-... 1 First of all, the expectation in question cannot be equal to\bar{X}$, since that is a random variable. (Your present working confuses the estimator with the parameter of interest.) From the specified distribution, the expected value of a single observation is:$\begin{equation} \begin{aligned} \mathbb{E}(X_i) &= \int \limits_\mathbb{R} x f(x|\theta)... 1 The estimator is unbiased for\theta$. It wouldn't make sense to say it's unbiased for$\bar{X}$or anything related to it, because$\bar{X}$is itself a random variable. Proof of unbiasness as below:$E \bar{X} = E X_1 = \int_{\theta}^{\infty} xe^{-(x-\theta)}dx= \int_0^\infty (u+\theta)e^{-u}du \quad$(let$u = x-\theta$)$= \int_0^\infty ue^{-u}du ...