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6

Sure, we can do so. Such random variables simply have degenerate densities or probability mass functions, such that the probability of an outcome is 1, or the outcome we are interested in may have a probability of 0. Such degenerate cases can easily come up, e.g., when we are looking at conditional probabilities. Let's throw a standard six-sided die. What ...

6

Your outcome variable $y$ is very overdispersed relative to a binomial, so the binomial model is misspecified. From one example of fit2, modified to quasibinomial: > summary(glm(formula = y ~ x + probe0 + probe1 + probe2, family = "quasibinomial")) Call: glm(formula = y ~ x + probe0 + probe1 + probe2, family = "quasibinomial") ...

5

As you apply the quotient rule on $\frac{u}{v}$, you have identify $v$ to be $2(1+|x|)$ but you did not square the $2$. Notice that $$\int_{-\infty}^\infty \frac{1}{(1+|x|)^2}\, dx=2.$$ Upon squaring the $2$ as well in the denominator, you should obtain the pdf to be $\frac1{2(1+|x|)^2}$.

4

Is the new variable IID ? Define $Y = h(X)$, you ask whether $\mathbf{Y} = \{Y_i\}_{i=1}^n \stackrel{\mathrm{i.i.d}}{\sim} p_Y$ if we have that $\mathbf{X} = \{X_i\}_{i=1}^n \stackrel{\mathrm{i.i.d}}{\sim} p_X$. Let's tackle a "simple" case, in which $h$ is invertible (and thus $h^{-1}$ exists). The CDF of $Y$ is F_Y(y) = P[Y\leq y] = P[\,h(X) \...

2

Let's simplify the notation and the problem a little. First, since the sample count is even, let's write it as $2n.$ Second, since $\theta$ appears as a location parameter, it suffices to solve the problem for a convenient value such as $\theta=0$ and then shift the result for general $\theta.$ Third, to avoid the fraction, let's find the distribution of ...

2

Per Wikipedia: If $X\sim\chi^{2}(\nu)$ and $c>0$, then $cX\sim\Gamma(k = \nu/2, \theta = 2c)$. Here, $\Gamma$ denotes the gamma distribution with $k$ and $\theta$ being the shape and scale, respectively. In your case, we have $2X\sim\Gamma(3/2, 4)$.

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