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## New answers tagged exponential-distribution

3

The joint likelihood is $$\mathcal L (\theta \mid \boldsymbol x) = \theta^{-n} e^{-n(\bar x - \theta)/\theta} \mathbb 1(x_{(1)} \ge \theta).$$ Since the unique critical point of the unrestricted likelihood occurs for $$0 = \frac{d}{d\theta} \left[\log \mathcal L \right] = -\frac{n}{\theta} + \frac{n\bar x}{\theta^2} = \frac{n}{\theta^2}(\bar x - \theta)$$ ...

0

I did a monte carlo simulation to model the problem because I didn't know how to do it analytically. Graphs at the bottom. Python code: https://github.com/jschaf/cellarch The simulation works as follows: Partition NUM_MACHINES into NUM_PARTITIONS separate groups. This simulation uses partition to refer to separate groups instead of cell used by the math ...

0

Here's an unconventional but for me more insightful way to do this: If $X_1, X_2, X_3, \cdots \stackrel{\text{iid}}{\sim} \text{Exp}(\lambda)$ then denote $S_k = X_1 + \cdots + X_k$, $k \geq 1$ to be the sequence of partial sums. I claim that for any fixed $n \geq 1$, the conditional distribution $(S_1, \cdots, S_n) | S_{n+1} = s$ is equal in distribution ...

2

Let $f_X$ be the density function of $X$. We have (by memorylessness) for all real $x,a$ \begin{equation}\Bbb P(X>x\mid X>a)=\Bbb P(X>x-a).\end{equation} Thus, by the Lebesgue Differentiation Theorem, $f_{X\mid X>a}(x)=f_X(x-a)$ for all $a$ and Lebesgue almost all $x$. It follows that (taken from the first answer on Conditional expectation of ...

1

HINT The first one is $$\begin{split} \mathbb{P}[U > 0.32\mid X > 0.19] &= \mathbb{P}[\min\{X,Y\} > 0.32\mid X > 0.19] \\ &= \mathbb{P}[X > 0.32, Y > 0.32\mid X > 0.19] \end{split}$$ Now use independence and compute. Similarly the second one...

-1

For part (a), note that $X_i-\theta$ are i.i.d $\mathsf{Exp}$ with mean $\sigma$. In other words, $\frac{2}{\sigma}(X_i-\theta)$ are i.i.d $\mathsf{Exp}$ with mean $2$, i.e. a $\chi^2_2$ variable. Therefore the correct pivot is $$T_1(\mathbf X,\theta,\sigma)=\frac{2}{\sigma}\sum_{i=1}^n (X_i-\theta)\sim \chi^2_{2n}$$ You can get a $100(1-\alpha)\%$ ...

1

Let $T_{1},T_{2},\ldots$ be a sequence of independent exponential random variables with parameter $\lambda=1/3$. Let $S_{n}\equiv T_{1}+\cdots+T_{n}$. By the memoryless property of the exponential distribution, $S_{n}$ is the time at which the $n$-th tanker arrives. Note also that $S_{n}$ has a Gamma distribution with parameters $n$ and $\lambda$.

2

You mention "amnesia" in the title of the post. This is a hint to use the memorylessness property of the exponential distribution. Solution sketch: Conditioned on the event $B$, the distribution of the remaining life time (after you walk in) of each bulb that is still lit when you walk in is also exponentially distributed with parameter $1$. Thus, the "...

1

A useful fact about MLEs you can prove with the chain rule is that $\widehat{f(\lambda)}=f(\widehat{\lambda})$. Therefore, $\widehat{\frac{1}{\lambda}}=\frac{1}{n}\sum_{i=1}^ny_i$.

1

Yes it's true. A more classical approach is to show that if $S_i\sim\exp(\lambda_i)$ then $$Ee^{-\xi}=\prod_{i=1}^\infty \frac{1}{1+\frac{1}{\lambda_i}}=\frac{1}{\prod_{i=1}^\infty\left(1+\frac{1}{\lambda_i}\right)}$$ and observe that the denominator is infinite if and only if $E\xi=\sum_{i=1}^\infty \frac{1}{\lambda_i}=\infty$ (see below why). In this case ...

1

Indeed $\overline X-\theta$ is a valid pivot for $\theta$ when $\beta$ is known. All you have to do now is find $a,b$ such that $a<\overline X-\theta<b\iff \theta\in(\overline X-b,\overline X-a)$ with the desired confidence coefficient $P_{\theta}\left[\theta\in(\overline X-b,\overline X-a)\right]$ (for a two-sided interval). This is easily done using ...

1

Hope this example helps. Let's say you use two identical alarm clocks to set an alarm to wake you up. Each alarm goes off at a random time which follows the same probability distribution. Let's call these times for the first and the second clock $X_1$ and $X_2$. Now $X_1$ and $X_2$ are i.i.d., and $\mathbb P(X_1 = \min (X_1, X_2))$ is equivalent to asking "...

1

If you have some probability function $P(t)$ for $t\geq0$, then the mean value of $t$ is given by $$\left<t\right>=\int_0^\infty tP(t)\text{d}t$$ Because you are shifting the start time, you need to adjust the normalization constant; you find this by solving $$A\int_T^\infty \text{e}^{-\lambda t}dt = 1$$ where $A$ is your new normalization ...

5

The mistake is that the waiting time is not distributed in the way that you have proposed, because the waiting time is the lesser of the service times of $B$ and $C$. This is because $A$ takes the first available opening, so whoever finishes first, $A$ chooses that clerk. That is to say, if $W_A$ is the waiting time of $A$ and $S_B$, $S_C$ are the service ...

0

Hint for part 2: $$E(g(X)) = \int_{-\infty}^{\infty} g(x) f_{X}(x) dx$$ In particular, $$E(-\log(U)) = \int_{0}^{1} -log(x) dx$$

1

First, you should be careful with writing $\mathbb{P}(X=x)$ with continuous RVs $X$ that have no point masses, because $\mathbb{P}(X=x)=0$ for all reals $x$ and all such continuous RVs, $X$. If you want to use the law of total probability in such a case, you apply it to the probability density function: $f(x)=\int f(x|y)f_Y(y)\mathrm{d}y$, or the cumulative ...

2

In (c), $\lambda$ is known so your complete sufficient statistic is $X_{(1)}$. By Lehmann-Scheffe, an unbiased estimator of $P(X_1\le t)$ which is a function of $X_{(1)}$ will be the UMVUE of $P(X_1\ge t)$. Let $g(\cdot)$ be that function. Now pdf of $X_{(1)}$ is $$f_{X_{(1)}}(z)=\frac{n}{\lambda}e^{-n(z-\theta)/\lambda}\mathbf1_{z>\theta}\quad, \,\... 0 Hint: In the proof of deriving Exponential from Poisson, you can observe that the Poisson distribution with n occurrence is equal to the distance between two tail probability of Erlang distribution with parameters (n,\lambda) and (n+1,\lambda) Some further algebraic manipulation should give you the proof 0 The random variable Y has a Laplace distribution. See Wikipedia at the start for a definition of the family of Laplace distributions and relevant bullets 3 and 10 under "Related Distributions." Your random variable Z is a modification of a Bernoulli; Bernoulli takes values 0 and 1.$$f_Y(y) = .5e^{-|y|}$$and F_Y(y) = .5e^y, for y<0 and F_Y(... 3 If y \geq 0 then:$$\begin{align}\def\P{\operatorname{P}} \P(Y \leq y)&=\P(X \leq y,Z=1)+\P(-X \leq y, Z=-1)\\[1ex]&=\P(X \leq y)\P(Z=1)+\P(X\geq -y)\P(Z=-1)&\star\\[1ex] &=\tfrac 1 2 (1-e^{-y})+\tfrac 1 2\\[1ex]&=(1-\tfrac 12e^{-y})\end{align} $\star$ assuming that $X$ and $Z$ are independent. Can you now handle the case $y <0$?

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