Let $Y_1, \ldots, Y_n$ be a random sample from the pdf $$f(y\mid\theta) = \theta y^{\theta - 1}, 0 &lt; y &lt; 1, \theta &gt; 0 .$$ show that $$E\left(\frac{1}{2\theta\sum_{i = 1}^n W_i}\right) = \frac{1}{2(n-1)}$$  I have showed that $\sum_{i = 1}^n -\ln(Y_i)$ is sufficient for $\theta$, and if $W_i = -\ln(Y_i)$, I have showed that $W_i$ has an exponential distribution with mean $1/\theta$. I have also showed that $2\theta \sum_{i=1}^n W_i $ has a $\chi^2$ distribution with $2n$ degrees of freedom.  My first question can we treat $2\theta\sum_{i=1}^n W_i$ as just 1/y? If so, I got to the following step $$\int_0^\infty \frac{1}{y} f(y) dy = \frac{1}{\Gamma(n)} \frac{1}{2} \int _0^\infty u^{n-2} e^{-u} du$$... I am wondering how to solve $\int _0^\infty u^{n-2} e^{-u} du?$ I know it is going to be $\Gamma(n-1)$ but what is the process for finding it?
122532
Let $Y_1, \ldots, Y_n$ be a random sample from the pdf $$f(y\mid\theta) = \theta y^{\theta - 1}, 0 &lt; y &lt; 1, \theta &gt; 0 .$$ show that $$E\left(\frac{1}{2\theta\sum_{i = 1}^n W_i}\right) = \frac{1}{2(n-1)}$$  I have showed that $\sum_{i = 1}^n -\ln(Y_i)$ is sufficient for $\theta$, and if $W_i = -\ln(Y_i)$, I have showed that $W_i$ has an exponential distribution with mean $1/\theta$. I have also showed that $2\theta \sum_{i=1}^n W_i $ has a $\chi^2$ distribution with $2n$ degrees of freedom.  I am wondering how to show it?
68054
Let $X_1,X_2,X_3 ...X_n$ be a random sample from Bernoulli   distribution with parameter $\theta$.Find UMVUE of $\theta(1-\theta)$.   I know that $T=\sum_{i=1}^{n}X_i$ is complete sufficient statistic for our paramenter $\theta$. I am trying to find out function of T which is a unique unbiased estimator of $\theta(1-\theta)$. Now $T\sim Bin(n,\theta)$  $E(T^2)-E(T)^2=V(T)$  $n\theta(1-\theta)+n^2\theta^2-n^2\theta^2=n\theta(1-\theta)$  $\implies\dfrac{1}{n}(T^2-\bar{T}^2)$ is umvue of $\theta(1-\theta)$     If I have sample size $n=10$ with observations $1,1,1,1,1,0,0,0,0,0$   obtain the value of this estimator.   Now I am stuck at this point that is $T^2$ is $T^2=\sum_{i=1}^{n}X_i^2$ or $T^2=(\sum_{i=1}^{n}X_i)^2$. Can someone help me and tell at what point I am doing things wrongly?
119742
Let $X_1, \dots, X_n$ be i.i.d. with the PDF given by $f(x \mid \theta) = 2\theta^{-1} xe^{-\frac{x^2}{\theta}}$ for $x &gt; 0$. My task is to find the UMVUE for $\theta$, and I’m given the following hint: “$U(X)=\sum_{i=1}^nX_i^2$ is sufficient for $\theta$, and $U/\theta$ is a multiple of $\chi_{2n}^2$”. It’s my understanding that to find the UMVUE, you need to derive $E[T \mid U = u]$ where $T$ is an unbiased estimator and $U$ is a complete sufficient statistic. However, I’m not sure how to make sense of the given hint. Where does the Chi-squared come from?
102954
Let $X$ ~ $U(0,2)$ and $Y$ ~ $U(-10,10)$ be two independent random variables with the given distributions. What is the distribution of $V=XY$?   I have tried convolution, knowing that   $$h(v) = \int_{y=-\infty}^{y=+\infty}\frac{1}{y}f_Y(y) f_X\left (\frac{v}{y} \right ) dy$$  We also know that $f_Y(y) = \frac{1}{20}$,    $$h(v)= \frac{1}{20} \int_{y=-10}^{y=10} \frac{1}{y}\cdot \frac{1}{2}dy$$ $$h(v)=\frac{1}{40}\int_{y=-10}^{y=10} \frac{1}{y}dy$$  Something tells me, there is something weird here since it is discontinuous at 0. Please help.
62728
Let $x_1, x_2... x_n$ be a random sample from a distribution with pdf:  $$f(x;\mu,\sigma)=\frac1{\sigma}\exp\left({-\frac{x-\mu}{\sigma}}\right)\,,-\infty&lt;\mu&lt;\infty;\, \sigma&gt;0;\, x\ge\mu$$  How do I find the MLE for the parameters if both parameters are unknown?  I tried using the usual MLE with likelihood function:  $$L(\mu,\sigma|x_1...x_n)=\frac{1}{\sigma^n}\exp\left({-\frac{\sum{x_i}-n\mu}{\sigma}}\right)$$ But the derivative of this with respect of $\mu$ is a dead end.   I do know that if $\sigma$ is known, the MLE for $\mu$ is $\frac{\sum{x_i}}{n}$ and if $\mu$ is known, the MLE for $\sigma$ is $\frac{\sum{x_i}-n\mu}{n}$.   Do these help? What should be the approach?
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?
85179
For two independent, discrete, uniformly distributed random variables $X$ and $Y$, I wish to obtain the distribution of the sum $Z=X+Y$. I have the densities: $$f_X(x)=\left\{\begin{matrix} \dfrac{1}{m+1},\;x\in [0,m]\\0,\;\text{otherwise}\end{matrix}\right.$$ $$f_Y(y)=\left\{\begin{matrix}\dfrac{1}{n+1},y\in [0,n]\\0,\;\text{otherwise}\end{matrix}\right.$$ $$f_Z(z)=\sum_{x=-\infty}^\infty f_X(x)f_Y(z-x)$$  I know that $f_Z(z)$ should look like a trapezoidal distribution. However, without knowing this, is it possible to calculate that distribution in a convenient way symbolically?   I'm finding it tedious to keep track of summation limits and such. For instance, separate cases for $m&gt;n$ and $m&lt;n$. I was wondering if there is a natural or intelligent notation for this, since this problem seems to be just so symmetric w.r.t $\,X$ and $Y$.  To make the question concrete: what is the most natural, symbolic/non-graphical way to solve for $f_Z(z)$? (I don't mind using another convolution formula if there is one, by the way)  Any help would be appreciated. Thanks.
65225
Let $X_i$ for $i=1,2,...$ be a sequence of i.i.d exponential random variables with common parameter $\lambda$. Let $N$ be a geometric random variable with parameter $p$ that is independent of the sequence $X_i$. What is the pdf of the random variable $Y=\Sigma_{i=1}^N X_i$.
97436
If I have the expression below, how can I use y~N($\mu$, $\Sigma$) to show it is a $\chi^2$ distribution with ($_n-1$) degrees of freedom?  ($n$-1)($s^2$)/$\sigma^2$ how is this ~$\chi^2$ $_n$-$_1$?     I recognize this contains SSE. I know that I should show the square root produces a distribution of $N$($\mu$, $I$) . I am lost at this point. I am thinking perhaps to substitute quadratic form into the original expression, then see what cancels? Or is this totally wrong?
11014
Let$ X_1, ...,X_n$ be a random sample feom a distribution $Geometric(\theta)$ for $0&lt;\theta&lt;1$. I.e,  $$p_{\theta}(x)=\theta(1-\theta)^{x-1} I_{\{1,2,...\}}(x)$$  Find the unbiased estimator with minimum variance for $g(\theta)=\frac{1}{\theta}$  My attempt:  Since the geometric distribution is from the exponential family, the statistics $$\sum X_i $$ is complete and sufficient for $ \theta$. Also, if $$T(X)=X_1$$ is an estimator for $g(\theta)$, it is unbiased. Therefore, by the Rao-Blackwell theorem and Lehmann-Scheffé Theorem, $$W(X) = E[X_1|\sum X_i]$$ is the estimator we are looking for.  We have the following:  $W(X) = \sum_{i=1}^t i\, P(X_1=i|\sum X_i =t) = \sum_{i=1}^t i\, \frac{P(\sum_{i \geq 2} X_i  =t-i)P(X_1=i)}{P(\sum_{i \geq 1}X_i =t)}$  Since the variables are iid geometric, the sums distributions are both negative binomials. But i am having troubles tosimplify the binomial coefficients and give a final answer with a better form, if it is possible.I wpuld be glad if I could get some help.  Thanks!  Edit: I dont think you guys understand my doubt: Ithink I made all the correct steps, maybe only forgot some indicator function. Here is what I did:  $$...=\sum_{i=1}^ti\frac{\binom{t-i-1}{n-2}\theta^{n-i}(1-\theta)^{t-i-n+1} \theta(1-\theta)^{i-1}}{\binom{t-1}{n-1}\theta^n(1-\theta)^{t-n}}=\sum_{i=1}^t i \frac{\binom{t-i-1}{n-2}}{\binom{t-1}{n-1}}$$  As i said, I am having troubles to simplify this and with the somatory index
42239
Suppose $X_1,\ldots,X_n$ are a random sample of $U(-\theta,\theta)$ with $\theta&gt;0$. How can find $$ E\left(|X_1|\ \big|\max |X_i| \right) $$ that $1\leq i\leq n$?
61165
For the pdf $f_{\theta}(x)=e^{-(x-\theta)} , x \ge \theta$, find a most powerful test of size $\alpha$, using Neyman Pearson Lemma to test $\theta=\theta_{0}$ against $\theta=\theta_1(&gt; \theta_0)$, based on a sample of size $n$.  I am facing difficulty as the parameter here is range dependent  However, if $X_{(1)}&gt;\theta_1$, then $f_1(x)&gt;\lambda f_0(x)$ if $e^{n(\theta_1- \theta_0)}&gt; \lambda$ would mean rejection of null hypothesis. But how will I make this test a size $\alpha$ test? The ratio is coming to be constant. Please help!
123783
There is something that I am doing wrong in the exercise below and I would appreciate some help figuring it out.  Let $X_1, X_2, \ldots, X_5$ be a random sample from a $\Gamma \left(3,3 \right)$ distribution . Define $W=\sum_{i=1}^5 X_i $.    Obtain the distribution of $W$ and of $2W/3$ Find $c_1$ and $c_2$ such that $P \left( c_1 &lt; W &lt;c_2 \right)=0.95 $     Since $W$ is the sum of iid gamma variables, $W \sim \Gamma \left(15,3 \right)$ and now a straightforward transformation argument shows that $Y=\frac{2}{3} W \sim \chi^2 \left(30 \right) $. This completes question 1.  Now for question 2, my thoughts were to transform these probabilities using the result from question 1, namely that $\frac{2}{3} W \sim \chi^2 \left(30 \right) $. Hence, unless I have made a horrible mistake here:  $$P \left( \frac{2}{3} c_1 &lt; Y &lt; \frac{2}{3} c_2 \right)=0.95 $$  We can find the constants from a chisquare table for $30$ degrees of freedom. In my case, they are $16.791$ for the lower bound and $46.979$ for the upper one.   Then if we equate $\frac{2}{3}c_1=16.791$ and solve for $c_1$ we should be able to find the constant for the above gamma confidence interval, right?   The problem is that according to $R$, $P \left( W&lt;c_1 \right) =1 $, instead of  the desired $0.025$ which means I have gone wrong somewhere. I do not think it is in question though $1$ as I have verified the result many times.   Thank you.
29766
Help me to solve this problem please..  Let $Y_{(1)}, Y_{(2)}, Y_{(3)}, Y_{(4)}, Y_{(5)}$ denote the order statistics of a random sample of size 5 from a distribution having p.d.f. $f(y) = e^{(-y)}, 0 &lt; y &lt; \infty$, zero elsewhere. Show that $Z_1 = Y_{(2)}$ and $Z_2 = Y_{(4)} − Y_{(2)}$  are independent. Hint: First find the joint p.d.f. of $Y_{(2)}$  and  $Y_{(4)}$.
13365
I am doing some independent study in asymptotic statistics and point estimation and am aware that you can get from log transformations of uniform random variables (exponential) all the way up to chi-squared distributions but I came across an intriguing question and I'm not sure how to handle it. I was thinking CLT or Delta Method but am unsure of myself. The thing I am trying to sort out is:  Given $X_1,\ldots X_n\overset{iid}{\sim}Uniform(0,\theta)$ and $Z_n=\left(\prod_{i=1}^nX_i\right)^{1/n}$ what is an example of a function of $Z_n$, i.e. $g(Z_n)$, so that $g(Z_n)\overset{d}{\rightarrow}\chi_{(1)}^2$, where $\overset{d}{\longrightarrow}$ denotes convergence in distribution and $\chi_{(1)}^2$ is a chi-squared random variable with 1 degree of freedom?  Maybe something like $g(h(Z_n))$ where $h(Z_n)\overset{d}{\rightarrow}Normal(0,1)$ and $g(t)=t^2$? Any help is greatly appreciated.  I have tried:  Let $Y_1=X_i/\theta$ I get $Y\sim Uniform(0,1)$ and so $-\log(Y)\sim Exponential(1)$. From here I know $-2\log(Y)$ is $\chi_{(2)}^2$ but I don't see how to relate $-\log(Y)$ back to a function of $Z_n$ that is $\chi_{(1)}^2$
66493
Given that $X$ and $Y$ are independent variables and that $X$ is exponential with a mean of $2$ and $Y$ is exponential with a mean of $1$. Find $P(X&lt;Y)$?   From the information it can be concluded that $X \sim \operatorname{Exp}(\frac{1}{2})$ and $Y \sim \operatorname{Exp}(1)$  Hence, $f(x,y) = \frac{1}{2}e^{-\frac{1}{2}x} e^{-y} $  Therefore, $P(X&lt;Y) = \int_{0}^{\infty} \int_{0}^{y} \frac{1}{2}e^{-\frac{1}{2}x} e^{-y} \,\,dx dy = \frac{1}{3}$   Is this the correct way to solve this problem?
59756
Say we have $f(x , \theta) = \frac{1}{2}e^{-|x-\theta|}$ Lets assume for simplicity, we only have 1 sample. We find that the log-likelihood for this distribution is:  $$ l(\theta , x) = -log(2) + (\theta - x) I_{(x &lt; \theta)}+(x-\theta)I_{(x \geq \theta)}. $$  Now in general, Call the score function:  $$ S(\theta , X) = Dl = \frac{Dg_{\theta}}{g_{\theta}} $$ by noticing the log-likelihood is expressed as :  $$ l(\theta, X) = log (g_{\theta}) $$  Where g is the likelihood and assume $g$ is 3 times differentiable. Furthermore, we find :  $$ DS = \frac{D^2g_\theta}{g_\theta}- SS'. $$ Which finally allows us to obtain that :  $$ var_\theta S = -E_\theta DS $$ by noticing that $E_\theta S = 0$  Now in this case:  $$ Dl = \frac{dl}{d\theta} = I_{(x &lt; \theta)} - I_{(x \geq \theta)} $$  So indeed  $$ E_\theta S = P{(x &lt; \theta)} - P{(x \geq \theta)} = 0 $$ And then $\frac{dS}{d\theta} = 0$ so $-E_\theta \frac{dS}{d\theta} = 0$  Now computing  $$ ES^2 = E( [I_{(x &lt; \theta)} - I_{(x \geq \theta)}]^2) = E(I_{(x &lt; \theta)} + I_{(x \geq \theta)}) = 1. $$ This does not match the case above, however here do not have differentiability. Is what I did correct, or plain fiction?