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An alternative viewpoint is that $X$ has a mixture density that consists of an equally-weighted sum of the density of an exponential random variable with parameter $2$ (the $e^{-2x}\mathbf 1_{\{x>0\} …

You need to calculate $E[e^{sU}e^{tV}] = E[e^{sX_1 + tX_1^2 + sX_2 + tX_2^2}] = E[e^{sX_1 + tX_1^2}]E[e^{sX_2 + tX_2^2}]$. So, set up the two integrals using the law of the unconscious statistician …

Since $$\operatorname{cov}(U,V)=a^2\operatorname{cov}(Z,Y)+ab\operatorname{cov}(Z,X)+ab\operatorname{cov}(Y,Y)+b^2\operatorname{cov}(Y,X)=ab\operatorname{var}(Y)\geq 0,$$ $U$ and $V$ are dependent if …

Let $f(\cdot)$ denote the density function of a continuous positive random variable and define $$g(x,y) = \begin{cases}\displaystyle \frac{f(x+y)}{x+y}, & x, y > 0,\\ 0, & \text{otherwise.}\end{cases …

The bivariate random variable $(X,\ln X)$ does not have a joint density and so you cannot use a double integral as you have indicated in your question. Instead, the law of the unconscious statisticia …

Hint: The integrand $e^{(t-1)x}$ has value $1$ if $t=1$ and is an increasing function of $x$ if $t > 1$. What do you suppose is the value of the integral when $t \geq 1$? In particular, is the value a …

Since you mention waiting times. Assume that $\alpha$ is an integer, and that $X_1$ is the time of occurrence of (equivalently, the waiting time for) the $\alpha$-th arrival after $t = 0$ in a Poisso …

Since the OP seems to be having difficulty with the various hints in the comments and the other answers, here is a heuristic method that yields the right answer in this instance. \begin{align} E[\exp( …

Assuming that $K$ takes on values $1,2,3,\ldots$ with $P\{X=k\} = (1-\beta)^{k-1}\beta$, $k > 0$, and not $(1-\beta)^{k}\beta$ as the problem states, then $P\{K > k\} = (1-\beta)^{k}$, either by rec …

I tried finding it with first principles and transformation formula but the calculation were so messy that I gave up. Well, first principles and the transformation formula say that with $Y = (X+a …