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Yes, you're just reparametrizing. Let $\gamma=\mu^2$ and let $L^*$ be the likelihood using $\gamma$ and $L$ the likelihood using $\mu$. Then $L^*(\gamma)=L(\sqrt{\gamma})$. Now just recall that the Cramer--Rao lower bound is the inverse of minus the expectation of the Hessian of the loglikelihood divided by $n$, which equals $\sigma^2$ if you're ...

2

The first integral is correct; the second is incorrect. You should write that if $1 < z \le \tfrac{3}{2}$, then $f(z-y) = 2$ if $z - \tfrac{1}{2} \le y \le 1$. This gives the correct density $f_{X+Y}(z) = 6-4z$ over $1 < z \le \tfrac{3}{2}$. A simpler way to do this calculation is to observe that $X^* = Y - \tfrac{1}{2} \sim {\rm ... 2 Hint: Use$f(X_1,X_2)=\int_0^1\int_0^1f(X_1,X_2|p_1,p_2)dp_1dp_2$and then you may want to take a look at Beta function: $$\int_0^1 p_1^{x_1}(1-p_1)^{n_1-x_1}dp_1=B(x_1+1,n_1-x_1+1)=\frac{x_1!(n_1-x_1)!}{(n_1+1)!}.$$ 1 Here is a hint: The density of a$X \sim {\rm Gamma}(\alpha,\beta)$random variable (shape, scale) is given by $$f_X(x) = \frac{x^{\alpha-1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)}, \quad x > 0.$$ We know that this function integrates to$1$over its support. Can you rewrite the integrand you obtained in your expectation to fit such a gamma density ... 1 $$L(p_1,p_2)={n_1\choose x_1}p_1^{x_1}(1-p_1)^{n_1-x_1}{n_2\choose x_2}p_2^{x_2}(1-p_2)^{n_2-x_2}$$ 0 The condition$np > 5$is not the condition, merely a rough estimate of what should be true in order for the normal distribution approximation to be "good enough". From Wikipedia: One rule is that both$x=np$and$n(1 − p)$must be greater than 5. However, the specific number varies from source to source, and depends on how good an approximation ... 1 Whether or not you consider the class of 8 students to be a population or a sample depends on the question you are interested in asking: "Is there a difference in the mean height among males versus females in that class only?" "Is there evidence to suggest that there is a difference in mean height among males versus females for all such classes in ... 1 You have assumed that$X$and$Y$are independent random variables and so I suppose that the actual problem you are solving is stating this somewhere, or it is taken from a section titled Independent Random Variables in some book. For a (continuous) random variable$Y$with density$f_Y(y)$, the density of$Z = aY$is $$f_Z(z) = \frac{1}{|a|}f_Y\left(\frac ... 2 Your answer is clearly wrong since the density does not integrate to 1 across the positive reals, as Dilip Sarwate. The way I would approach this would be to take your f(x,y)=\dfrac{e^{-y/4}}{8} and look at$$\Pr\left(\frac{Y}{X} \le u\right)=\int_{x=2}^{4} \int_{y=0}^{ux} \dfrac{e^{-y/4}}{8} dy\, dy= \int_{x=2}^{4}\left(\frac12- ... 1 If$x_1,\ldots,x_n$are the scores in$n$categories assigned to some school, pick some$p$from$(0,\infty)$and compute the total score as $$\bar x = \left(\frac{1}{n}\sum_{k=1}^n \textrm{sgn}(x_k)\sqrt[p]{|x_k|} \right)^p \text{ where }\textrm{sgn}(x) = \begin{cases} -1 & \text{if x < 0} \\ 0 & \text{if x=0} \\ 1 &\text{if x > ... 0 Consider the following. Assume that the first two of your variables are perfectly correlated and orthogonal to the third. If a user specifies (1/3,1/3,/1/3) as their weights, then they are saying that the third category should count as 1/3 of the total score REGARDLESS of how the first two categories may be correlated. Thus, correlation or ... 3 A standard way to compare two (sufficiently nice) functions f(x) and g(x) over the interval [a,b] is to use the inner product$$\left<f(x),g(x)\right>:=\int_a^b{f(x)g(x)\,\mathrm{d}x}$$from which we get$$||f(x)-g(x)||=\sqrt{\int_a^b{\left(f(x)-g(x)\right)^2\,\mathrm{d}x}}$$where you can think of ||f(x)-g(x)|| as being the "distance" between ... 1 The logistic and logit transformations have advantages such as: (as you identified) symmetric in a sense between (0,1) and the whole real line analytically tractable: integral and derivatives have closed forms a natural log-odds interpretation: x=\log\left(\frac{p}{1-p}\right) the logistic distribution being close to a normal distribution (in practice ... 1 If the variables are jointly normal, then their sum, denote it S_n will be a normal random variable, with mean zero and a variance, denote it simply by v_s. Then the random variable Z = S_n^2/v_s will follow a chi-square distribution with one degree of freedom. Then, the random variable W = v_sZ =S^2_n follows a Gamma distribution with shape ... 0 Hint: \Pr(\text{black ball is chosen from urn 1)}=\frac{4}{11} \Pr(\text{black ball is chosen from urn 2})=\frac{3}{8} Now you need to figure out what to do with these probabilities. 0 A sample mean of 24 is unlikely from a sample of twenty-five items with a population mean of 25 and population standard deviation of 1. So the sample casts doubt on the population parameters. But conditioned on the data given and ignoring issues such as finite populations, a good estimate of the variance of the sample conditioned on the population ... 0 Standard deviation is 0 because all the samples are completely random. 0 One has$$ \hat\beta_1 = \frac{\sum_{i=1}^n (y_i-\bar y)(x_i-\bar x)}{\sum_{i=1}^n (x_i - \bar x)^2} $$where \bar y = (y_1+\cdots+y_n)/n and \bar x = (x_1+\cdots+x_n)/n. This is nonlinear as a function of x_1,\ldots,x_n since there is division by a function of the xs and there is squaring. But it is linear as a function of y_1,\ldots,y_n. To ... 2 I've not worked out the details but I am sure it will go through. Calculate the moment generating function E[exp(s \sum X_i)]. Differentiate this enough times and you get your answer. Taking the covariance matrix to have compound symmetry then I get the generating function for \sum X_i to be exp(n(1+(n-1)\rho)\sigma^2 s^2 /2). This is at least right ... 0 The answer is affirmative. Every positive semidefinite matrix C can be orthogonally diagonalised as QD^2Q^T, where Q is a real orthogonal matrix and D is a nonnegative diagonal matrix. Let \mathbf{Z} be a random vector following the standard multivariate normal distribution N(0,I_n). It is straightforward to verify that C is the covariance ... 2 It is possible to find the joint distribution of X and X^2, but that is the hard way of tackling the problem. Informally, if we know that X is close to 0, then X^2 cannot be close to 1. We turn this observation into a formal argument. It is easy to verify that \Pr(0\lt X\lt 1/2)\ne 0. Also, \Pr(X^2\gt 1/2)\ne 0. But$$\Pr((0\lt X\lt ... 0 Since in a comment you say the tests should have equal weight, probably we should first scale the means and standard deviations, so they are all out of some common value. In your example, you could choose$100$for each, or the least common multiple of the denominators. We will assume that sample standard deviation was computed by dividing by the class ... 1 Note that$\mathbf x$and$\hat{\mathbf x}$are vectors indexed by the vertex set$V$. The assumption in (1) is that every entry of$\mathbf x$and$\hat{\mathbf x}$corresponding to some vertex in the clique$C$coïncide. Example: if$V=\{1,2,3,4,5\}$and$C=\{1,2\}$, condition (1) asks that some function$g$on$\Lambda^5$is such that ... 2 Your function$x\mapsto\dfrac{\exp(-\sqrt{x})}{\theta}$is not a probability density on$(0,\infty)$unless$\theta=2$. Here's a guess: Maybe you meant$\dfrac{\exp\left(-\sqrt{x/\theta\, {}}\right)}{2\theta}$. At any rate, suppose you have i.i.d. observations, and you want $$\mathbb E\left(\frac{\sqrt{X_1}}{2} \mid \sqrt{X_1}+\cdots+\sqrt{X_n}\right).$$ ... 0 There is no$X_C$here and, by definition, $$x_C=(x_v)_{v\in C}.$$ (This notation is used everyuwhere on the page linked to, and well before the paragraph reproduced in the question.) 0 Since the rolls are independent and a roll can result in each of the values from 1 to 6 with equal probability 1/6, E(X1) = 1.1/6 + 2.1/6 + ..... + 6.1/6 = 3.5 = E(X2). Thus, E(Y) = E(X1) + E(X2) = 7 and E(Z) = 4E(X1) - E(X2) = 10.5. Now, if you right down all possible values of YZ with their corresponding probabilities, you'll arrive at E(YZ) = 82.25.(**) ... 2 The easiest and most elegant way to find a formula for this is to show you an equivalent problem. This is the case of indistinguishable objects in distinguishable boxes. Rather than thinking that a Skittle is a particular color, let's have identical beads be placed in one of$5$boxes, and depending on which box it is in determines its color of the Skittle. ... 1 This is a multiset problem. So the count is $$\binom{colors + skittles - 1}{skittles}$$ So we have$5$colors and$2$skittles, so we get$\binom{5 + 2 - 1}{2} = \binom{6}{2} = 15$. 1 You can write the covariance as$E[XY] - E[X]E[Y]$. Then, use the definition of expectation:$E[XY] = \int_{\mathbb{R}^2} xy p_{X,Y}(x,y) dx dyE[X] = \int_{\mathbb{R}^2} x p_{X,Y}(x,y) dx dyE[Y] = \int_{\mathbb{R}^2} y p_{X,Y}(x,y) dx dy$or equivalently for the latter two,$p_X(x) = \int_{\mathbb{R}} p_{X,Y}(x,y) dyp_Y(y) = \int_{\mathbb{R}} ...

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You have to assume that the number of tetanus cases reported follows a Poisson distribution with $\lambda=4$. Now using the Poisson distribution you can find out the given probabilities easily, and also the standard deviation.

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If you integrate $p(x,y)$ over all $x$ and $y$, you should get $1$. This determines $c$. Then, if you integrate $y p(x,y)$ over all $x$ and $y$, you get the expected value of $Y$. Alternatively, note that the domain is a product set (in this case, a rectangle with its sides aligned with the x and y axes) and $p(x,y)$ factors into a function of $x$ and ...

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If two normal random variables are jointly normally distributed, then the pair $(X,Y)$ is distributed according to the multivariate normal distribution with mean $[0,0]^T$ and covariance $\Sigma=\left[\begin{array}{cc}1&0.4\\0.4&1\end{array}\right]$ (note that the correlation coefficient $\rho_{X,Y}$ happens to be equal to the covariance in this case ...

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If there is no intercept term, you shouldn't have a column of zeros. The matrix equation is $\begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix} = \beta \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}$, so $X = \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}$.

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What you might be erroneously computing, with $f(x,y)=\frac23(x+2y)$: $$\int_0^1\int_0^{1/3+y}f(x,y)\mathrm dx\mathrm dy.$$ What you should be computing: $$\int_0^1\int_0^{\min\{1,1/3+y\}}f(x,y)\mathrm dx\mathrm dy,$$ that is, $$\int_0^{2/3}\int_0^{1/3+y}f(x,y)\mathrm dx\mathrm dy+ \int_{2/3}^1\int_0^1f(x,y)\mathrm dx\mathrm dy.$$

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$$p_{X|Y}(x|y) = \frac{p_{X,Y}(x,y)}{p_Y(y)}$$ by Bayes rule (for $y$ such that $p_Y(y) \neq 0$) Now, note $p_Y(y) = \int_{\mathbb{R}} p_{X,Y}(x,y) dx$ (this is the y marginal of the distribution $p_{X,Y}$)

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Yes. Let $X_1,\ldots, X_5$ be independent and $P(X_i=0)=P(X_i=1)=\frac12$. Now let $X=X_1$ $Y=(\max\{X_2,X_3\}=X_1)$ $Z=(\max\{X_4,X_6\}=X_1)$ If $X=0$, the latter two simplify to $X_2=X_3=0$ and $X_4=X_5=0$, which are clearly independant. If $X=1$, they simplify to $X_2=1\lor X_3=1$ and $X_4=1\lor X_5=1$, which are clearly independant. However, $Y,Z$ ...

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I think this question was asked the other day, so this might be a duplicate. In any case, you must observe that $$Y \sim \chi^2_4$$ see here. UPDATE As requested in the comments, here is some more detail: using the formulae found in the link above, you see the density is $$f_Y(y) = \frac{1}{4} y e^{-y/2}, \quad y \geq 0$$ and the distribution ...

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You may be interested in the idea of a sufficient statistic. What the commenters above are pointing out is that there are very few distributions for which just the mean is a sufficient statistic. One notable example which comes to mind is a Poisson distribution - the variance is equal to the expectation, so you could determine the entire distribution just ...

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Actually, I don't need the support of Y and Z at all. "f($X_1,X_2$)=1 over their support" means that f($X_1,X_2$)=$\frac{1}{(10-4)(1-5)}$=$\frac{1}{30}$, so f(Y,Z)=$\frac{\sqrt{Z/Y}}{2\sqrt{YZ}}$*$\frac{1}{30}$=$\frac{\sqrt{Z/Y}}{60\sqrt{YZ}}$ Thank you, @Tunk-Fey for the help.

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The first part is right, the second part you have to integrate $p_{X,Y}$ over the small square. If X and Y are independent then $p_{X,Y}=p_{X} \times p_{Y}$ then what you say would be OK.

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Hint: If $Z_1,\ldots,Z_k$ are independent, standard normal random variables, then the sum of their squares, $$Q=\sum_{i=1}^k Z_i^2,$$ is distributed according to the chi-squared distribution with $k$ degrees of freedom.

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The covariance of $X$ and $Y$ is $E(XY)-E(X)E(Y)$. (The formal definition of covariance is $E((X-E(X))(Y-E(Y)))$, but that is usually, and in this case, harder to work with.) To find $E(XY)$, find the sum $\sum_{(x,y)} xy\Pr(X=x\land Y=y)$. There will be $9$ terms to add up, really only $7$, since $2$ of the terms are $0$, A typical term like the one ...

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This explicit computation may help: \begin{align} P[X+Y > 6] &= P[X +Y > 6, X=2] + P[X +Y > 6, X=3] \\ &= P[Y = 5, X = 2] + P[ Y \geq 4, X=3 ] \\ &= P[Y = 5]P[ X = 2] + P[ Y \geq 4]P[ X=3 ] \\ &= \left(\frac{1}{5} \right)\left(\frac{7}{10} \right) + \left(\frac{3}{5} \right)\left(\frac{3}{10} \right) \\ &= \frac{8}{25}. ...

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One time in six you get four dollars profit. Five times in six you lose one dollar. That is the probability distribution. The expected value just multiplies the chance by the value of the outcome and adds them up. $(\frac 16)\cdot 4 + (\frac 56)\cdot (-1)=?$

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Standard Deviation = sqrt(250*0.04*0.96) So about 3 in your case. Hope this helped!

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The number of defective toys is a binomial $(250,0.04)$ random variable - each toy is treated as an independent Bernoulli trial where the "success" is voicebox defect. It suffices to use the standard formula for variance of a binomial variable.

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The marginal in $x$ is $$\int_{x^2}^\infty 2xe^{-y} dy = 2x \left. (- e^{-y}) \right|_{x^2}^\infty = 2xe^{-x^2}, \quad x > 0.$$ The marginal in $y$ is $$\int_0^{\sqrt{y}} 2xe^{-y} dx = e^{-y} \left. (x^2) \right|^{\sqrt{y}}_0 = ye^{-y}, \quad y > 0.$$

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We will assume that by "difference does not exceed $0.5$ inch," what is meant is that the absolute value of the difference does not exceed $0.5$ inch. Let the heights of the people in the sample be $X_1,X_2,\dots,X_{100}$. Then the sample mean is the random variable $Y$, where $$Y=\frac{1}{100}(X_1+X_2+\cdots+X_{100}).$$ Let the population mean be $mu$. ...

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Draw a picture of the part of the plane on which the joint density function "lives." This is the triangle with corners $(0,0)$, $(1,0)$, and $(1,2)$. The argument below cannot be understood without the picture. We have $F_Z(z)=\Pr(Z\le z)$. It is a little easier to look for $\Pr(Z\gt z)$. First suppose $z\ge 1$. Then $Z\gt z$ precisely if $Y\gt z$. This ...

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If $X_2$ is exponentially distributed with parameter $1$, then $P(X_2\gt x)=\mathrm e^{-x}$ for every positive $x$, hence $$P(X_2\gt \color{red}{\mathbf 2}X_1)=E(\mathrm e^{-\color{red}{\mathbf 2}X_1}).$$ Furthermore, for every nonnegative $\color{blue}{\mathbf c}$,  E(\mathrm e^{-\color{blue}{\mathbf c}X_1})=\int_0^\infty\mathrm ...

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