the density for an exponentially distributed RV X with parameter λ&gt;0 is given as: f(x)=(1/λ )exp(−x/λ) for non-negative x and 0 otherwise. The expected value of the maximum of two independent exponentially distributed random variables X  and Y  with parameters λ1=1 and λ2=2  is...?
105214
I want to find the expected value of $\text{max}\{X,Y\}$ where $X$ ist $\text{exp}(\lambda)$-distributed and $Y$ ist $\text{exp}(\eta)$-distributed. X and Y are independent. I figured out how to do this for the minimum of $n$ variables, but i struggle with doing it for 2 with the maximum.  (The context in which this was given is waiting for the later of two trains, with their arrival times being exp-distributed).  Thanks!
46087
Given that $X_1, X_2...$ are mutually independent random variables. For each $i$ with $1\le i \le n$   the variable $X_i$ is equal to either $0$ or $n+1$ $E(X_i)$ = $1$   also.. if $X_i$ is equal to either $0$ or $n+1$, doesn't that mean that all $X_i$ need to be $0$?  Thanks
9750
Let be $X_1,X_2,...,X_n$ be uniformly distributed random variables i.i.d.  a)  Find the cumulative function and the density of $m_n \text{ and } M_n$ , where $m_n=min(X_1,X_2,...,X_n)$ and $M_n=max(X_1,X_2,...,X_n)$.  b)  Let $Z_n=n(1-M_n)$. Show that $Z_n \xrightarrow{d} Z $, where $Z$ is a random variable with cumulative function $F_Z(z)=1-e^{-z}$
44565
I know that the parameter space has to be convex (that I've proved using Jensen's inequality). Therefore, since it has to be a subset of $\mathbb R$, it follows that it has to be an interval or a point. Assuming that it is not a point, is it possible that the parameter space $\mathcal N$ for one-parameter exponential family distribution in canonical form, i.e. one with the following density: \begin{equation} f_\theta(x) = C(\theta)h(x)e^{\theta T(x)} \end{equation} is a closed interval?
7221
Been crushing my head with this exercise. I know how to get the distribution of a ratio of exponential variables and of the sum of them, but i can't piece everything together.  The exercise goes as this:  If X,Y are independent exponentially distributed with beta = 1 (parameter of the exponential distribution = 1) then what is the distribution of X/(X+Y)  Any ideas?  Thanks a lot.
76097
I am trying to define the probability distribution of $Z$ such as $Z = X_1\cdot X_2$ where $X_1$ and $X_2$ are two independent and identically exponentially distributed variables.  $$P(X_1=x) = \lambda e^{-\lambda x}$$ $$P(X_2=x) = \lambda e^{-\lambda x}$$  I tried something like that….  $$P(Z=x) = P(X_1 = a) \cdot P(X_2 = x/a)$$  Is this first equation correct?  $$P(Z=x) = \lambda e^{-\lambda a} \cdot \lambda e^{-\lambda x/a} = \lambda^2 e^{-\lambda (a+x/a)}$$  I doesn't feel like I found the solution as the dummy variable $a$ remains in the final result. Can you please help me solving this problem?
117331
Suppose that $X$ and $Y$ are chosen randomly and independently according to the uniform distribution from $(0,1)$. Define $$ Z=\frac{\max(X,Y)}{\min(X,Y)}.$$ Compute the probability distribution function of $Z$. Can anyone give me some hints on how to proceed? I can only note that $\mathbb{P}[Z\geq 1]=1$ and $$F_Z(t)= \mathbb{P}[Z \leq t]=\mathbb{P}[X\leq Y, Y\leq tX]+\mathbb{P}[Y \leq X, x\leq tY]$$
59799
Let's say I have an outcome that is exponentially distributed, so $p(y|\lambda_i) = \lambda_i e^{-\lambda_i y}$. However, we also know that $\log{\lambda_i} = \beta_0+\beta_1x_i$. If we want to estimate the MLE's for $\beta_0$ and $\beta_1$, what's the best way to do it given data?  I was thinking of using the following code:  summary(glm(y~x,family=Gamma(link="log")))   Not 100% sure what the various arguments are going though.
86866
I have $X_1,X_2 ... X_n$ Gamma-distributed r.v with density:    First of all I showed, that Gamma distribution belongs to exponential family and can be represented in form     I found, that for Gamma distribution $B(\theta) = log(\theta)$ where $\theta$ is $\alpha/\beta$  Then we know, that moment generating function of the random variable $T(X)$ is given by $$M(t)=\exp\{B(\theta+t)-B(\theta)\}$$  Natural sufficient statistics for Gamma distribution is:    But I didn't understand what should I do next? Because Natural sufficient statistics is two dimensional it confuses me:  $M(T) = exp[log(\theta + T) - log(\theta)]$  and $T = [s,t]^\intercal$
70929
I am trying to do an excerise on likelihood functions and score vectors and I am having some trouble extending it to parameter vectors. The question is stated below. Suppose we have two independent samples: $X_{1}, ..., X_{n}$ are iid Exponential random variables with mean $θ_{1}$ and $Y_{1}, . . . , Y_{n}$ are iid Exponential random variables with mean $θ_{2}$. Find the likelihood and log likelihood functions for the parameter vector $θ = (θ_{1}, θ_{2})^{T}. $ In this case would the likelihood function be $$\prod{θ e^{-θ x_{i}} = θ ^ne^{-θ\sum x_{i}}}$$ (same as one with one parameter)? Or would it be the product of the two functions as shown below: $$\prod{θ_{1} e^{-θ_{1} x_{i}} θ_{2}e^{-θ_{2}y_{i}} = θ_{1} ^ne^{-θ_{1}\sum x_{i}}} θ_{2}^{n}e^{-θ_{2}\sum y_{i}}$$ Or would it be something different? Thanks!
17117
Consider $Y_1, \dots, Y_n$ independent from the Gamma distribution. For $y &gt; 0$: $$\begin{align} f(y \mid \alpha, \beta) &amp;= \dfrac{1}{\beta^{\alpha}\Gamma(\alpha)}y^{\alpha-1}e^{-y/\beta} \\ &amp;= \exp\left[-\log(\beta^{\alpha}\Gamma(\alpha))+(\alpha-1)\log(y)-\dfrac{y}{\beta} \right] \\ &amp;= \exp\left[y\left(\dfrac{-1}{\beta} \right)-\log(\beta^{\alpha}\Gamma(\alpha))+(\alpha-1)\log(y) \right] \end{align}$$ I define an exponential dispersion family as any distribution whose PMF/PDF is  $$f(y \mid \boldsymbol\theta) = \exp\left\{\phi[y\theta - b(\theta)] + c(y, \phi) \right\}\text{, } y \in \Omega$$ where $\Omega$ is in the support of a random variable $Y$ in the family.    Method 1:  Therefore, $Y_i$ is of the exponential dispersion family with (assuming $\alpha_i$ is known) \begin{align} \phi &amp;= 1 \\ \theta_i &amp;= \dfrac{-1}{\beta_i} \\ b(\theta_i) &amp;= \log[\beta_i^{\alpha_i}\Gamma(\alpha_i)] \\ &amp;= \log[(-\theta_i)^{-\alpha_i}\Gamma(\alpha_i)] \\ &amp;= -\alpha_i\log(-\theta_i) + \log \Gamma(\alpha_i) \\ c(\phi, y_i) &amp;= (\alpha_i-1)\log(y_i)\text{.} \end{align} It follows that \begin{equation} \mu_i = b^{\prime}(\theta_i) = \dfrac{-\alpha_i}{-\theta_i}(-1) = \dfrac{-\alpha_i}{\theta_i}\text{.} \end{equation} and our canonical link is   $$g(\mu_i) = \theta_i = -\dfrac{\alpha_i}{\mu_i}\tag{*}$$    Method 2:  Credit due to .  Let $\theta = -\dfrac{1}{\alpha\beta}$. Then rewrite $$\begin{align}f(y \mid \alpha, \beta) &amp;= \exp\left[y\left(\dfrac{-1}{\beta} \right)-\log(\beta^{\alpha}\Gamma(\alpha))+(\alpha-1)\log(y) \right] \\ &amp;= \exp\left[y\left(\dfrac{-1}{\beta} \right)-\alpha\log(\beta)-\log\Gamma(\alpha)+(\alpha-1)\log(y) \right] \\ &amp;= \exp\left[y\alpha\theta-\alpha\log(\theta^{-1})-\log\Gamma(\alpha)+(\alpha-1)\log(y) \right] \\ &amp;= \exp\left\{\alpha[y\theta-\log(\theta^{-1})]-\log\Gamma(\alpha)+(\alpha-1)\log(y) \right\}  \end{align}$$ in which case, since $\phi$ cannot vary depending on $i$, \begin{align} \phi &amp;= \alpha \\ \theta_i &amp;= -\dfrac{1}{\alpha\beta_i} \\ b(\theta_i) &amp;= \log(\theta_i^{-1}) = -\log(\theta_i)\\ c(\phi, y_i) &amp;= (\alpha-1)\log(y_i)-\log\Gamma(\alpha)\text{.} \end{align} In this case, $b^{\prime}(\theta_i) =-\dfrac{1}{\theta_i} = \mu_i$, which means that $$g(\mu_i) = \theta_i = -\dfrac{1}{\mu_i}\text{.}\tag{**}$$     cites (**) as the canonical link function. Why is (**) preferable over (*) for the canonical link function?  What makes this even more confusing is that  says that the canonical link function is $\dfrac{1}{\mu_i}$ (also mentioned in McCullagh and Nelder).     First main question: Are canonical link functions unique, when provided a random component?      Second main question: Is there only supposed to be one canonical link function for each random component? If so, what is wrong with the work above?
14597
How do I find the pdf of $\prod_{i=1}^n X_i$, where $X_is$ are independent uniform [0,1] random variables.  I know  X~U[0,1], -ln(x) is exponential(1). I also know the sum of two or more independent exponential random variable is gamma.   For $Y = \sum_{i=1}^n -ln(x_i)$, which is a gamma(n, 1), I found the pdf for Y is $$\int_0^\infty \frac{1}{\Gamma (n)} y^{n-1} e^{-y} dy$$.  Let $Z = e^Y$  I am trying to the pdf for Z, what I found is $$\int_1^\infty \frac{1}{\Gamma (n)} ln(z)^{n-1} \frac{1}{z^2} dz$$, which does not look right to me. Could someone check it?
32806
$X$ and $Y$ are independent and exponentially ditributed with parameters $a,b$, respectively. Calculate $P(X&gt;Y)$  1)$P(X&gt;Y)=1-F_{X-Y}(0)$. Then I try to calculate density  of $X-Y$ from characteristic functions but I cannot deduce distribution of $X-Y$ from this.  2)ok I can also calculate density of $Z=X-Y$ from transformation of RV formula and then marginalize  3) Is there any simpler way to calculate $P(X&gt;Y)$ than 2)?
87869
I have an answer but I'm not sure if I'm doing it right: $f(x)$ is the pdf $f(x)=2e^{-2x}$, then, if the random variable $Y=\ln(X)$, then we simply substitute:  $P(Y \leq y)$=  $P(2e^{-2X} \leq y)$  $P(-4X \leq y)$ since $y=ln(x)$  $P(X \leq - \dfrac {1} {4} y)$  This is the distribution function, right?  Thanks!
28011
Let $Z_i,Z_2,\ldots$ be IID Random Variables with density $f$. Suppose that $P(Z_i&gt;0)=1$ and that $\lambda=\lim_{x \to 0+} f(x)&gt;0$. How can I show that $X_n=n \times \min\{Z_i\}$ has a limiting exponential distribution with mean$1/\lambda$?  I know that the CDF of the first order Statistic is for $t&gt;0$  $$F_{min \{ Z_i \}} (t)= 1- \left[ 1-F(t) \right]^{n}$$  Convergence in distribution requires that $F_n (t) \to F(t)$ for some distribution function $F$ but I cannot quite take the limit of the above, with $t$ replaced by $x/n$, as the distribution is not known. I guess I have to insert the second piece of information somewhere but I do not recognise where.  Could you please give me a hint or two?  Thank you.
70903
Given the identically distributed and independant random variables $X_1,X_2,\ldots\sim\operatorname{Po}(\lambda)$ and $S_n=X_1+\ldots+X_n$ show with induction that   $$\Pr[S_n=k]=\frac{(n\lambda)^k}{k!}e^{-n\lambda}.$$   So far for $n=1$ via the definition of the density function for Poisson:$$\Pr[S_1=k]=\Pr[X_1=k]=\frac{e^{-\lambda}\lambda^k}{k!}$$ With $n=2$ and independency: $$\Pr[X_1+X_2=k]=\sum\limits_{n=0}^k\Pr[X_1=n]\Pr[X_2=k-n]=\sum\limits_{n=0}^k\frac{\lambda^n}{n!}e^{-\lambda}\frac{\lambda^{k-n}}{(k-n)!}e^{-\lambda}$$$$=\frac{1}{k!}e^{-2\lambda}\sum\limits_{n=0}^k\binom{k}{n}\lambda^n\lambda^{k-n}=\frac{(2\lambda)^k}{k!}e^{-2\lambda}$$ I assume now that with $S_n\sim\operatorname{Po}(\lambda)$ every $X_i\sim\operatorname{Po}(\lambda/n)$ and every Poisson variable can obviously split up, but how can I prove this?
57255
I have an exercise where X has an exponential distribution with rate 1, and Y|X=x is a poisson distribution with parameter x.   I can find the expectation and variance of Y using the law of total expectation and law of total variance - and I have gotten 1 and 2 respectively - I'm fairly sure this is correct.  Now I am trying to find the probability generating function of Y, but I'm struggling. I think I need to use the law of total expectation to help me here, but I'm not too sure how.   Any help is appreciated.