Let $Z\sim N(0,1)$ and $Y=Z^2$. Find the distribution of $Y$ using moment generating functions. \begin{align*} M_Y(t)&amp;=E(e^{Z^2t})\\ &amp;=\int_{-\infty}^\infty e^{z^2t}f_Z(z)\ dz\\ &amp;=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{z^2t}\cdot e^{-\frac{z^2}{2}}\ dz\\ &amp;=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{z^2(t-\frac12)}\ dz \end{align*} I'm not sure how to simplify this or find which distribution this is
25527
If $X\sim N(0,1)$, integrate to find the moment generating function of a random variable $X^2$ and identify the distribution of $X^2$ using the moment generating function. $$E\left[e^{tX^2}\right]=\int_{-\infty}^{\infty}{e^{tx^2}e^{-x^2}\over{\sqrt{2\pi}}}dx$$ which reduces to $$=\frac1{\sqrt{2\pi}} \int_{-\infty}^{\infty}{e^{tx^2}e^{-x^2}}dx =\frac1{\sqrt{2\pi}} \int_{-\infty}^{\infty}{e^{x^2(t-\frac12)}}dx$$ and thus I am stuck. I'm sure there must be some trick to this (like completely the square for the mgf of a standard normal variable X) but I can't figure out what it might be. Any hints?
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?
114802
I'm trying to wrap my head around linear transformations to random variables (with coefficients > 1).  Consider the two random and independent variables $X$ and $Y$ where:  $$X \sim \mathcal{N}(0,1)\qquad Y \sim \mathcal{N}(-1, 2)$$  I'm trying to determine the distribution of $Z = 2X + 3Y$.    The attempt so far:  I've defined the MGF of both $X$ and $Y$ as:  \begin{align*}M_X(s) &amp;= \exp(s \mu + \sigma^2 s^2 / 2) = \exp(s^2/2)\\ M_Y(s) &amp;= \exp(-s + s^2)\\ \end{align*}  Then the plan was to calculate $U$ as:  $Z = 2X + 3Y = 2 \exp(s^2/2) \cdot 3 \exp(-s + s^2) = 6 \exp(3s^2/2 - s)$  This lead me to a dead end and I'm not sure if this is the right approach, the "new" MGF doesn't exactly look like anything I know.  I've considered using a convolution to try and solve this but I am not 100% about that approach.   There was a mention on  about how to do it for the case of the coefficients = 1 and that there was a special case for coeffs > 1 (however it failed to expand on it).    I've done some simulations in MATLAB with random sampling from both of these Gaussians  and then applying the transformation. The results indicate the solution should be (assuming the code is correct):  $Z \sim \mathcal{N}(-3, 22)$
39125
If we know the moment generating functions (MGFs) of the random variables $X$ and $Y$ to be $M_{X}(s)$ and $M_{Y}(s)$, respectively. The MGF of the sum $X+Y$ will $M_{X}(s) \cdot M_{Y}(s)$.  So what is the MGF of the ratio distribution $\displaystyle\frac{X}{Y}$?
43705
I've run into a practice problem where $Y = \cos(\pi X)$, and $X$ is uniform on $[0, 1]$, and I'm supposed to prove that the PDF of $Y$ is $1/(\pi\sqrt{1 - y^2})$. I know that for derived distributions, you plug in the the equation for $Y$ in terms of $X$ into the PDF of $X$, integrate up to $y$, and then differentiate with respect to $y$. However, when I do this, I get the PDF of $Y = -1/(\pi\sqrt{1 - y^2})$.   I'm sure that the problem here is that there's some property of the cosine and its inverse that I don't know, so if someone could point out how to do this, that would be great, thanks!
116875
let $Z \sim N(0,1)$ and $W \sim \chi^2(s)$.  If $Z$ and $W$ are independently distributed then the variable $Y = \frac{Z}{\sqrt{W/s}}$ follows a $t$ distribution with degrees of freedom $s$.  I am looking for a proof of this fact, a reference is good enough if you do not want to write down the complete argument.
113561
If $U$ and $V$ are independent identically distributed standard normal, what is the distribution of their difference?  I will present my answer here.  I am hoping to know if I am right or wrong.  Using the method of moment generating functions, we have  \begin{align*} M_{U-V}(t)&amp;=E\left[e^{t(U-V)}\right]\\ &amp;=E\left[e^{tU}\right]E\left[e^{tV}\right]\\ &amp;=M_U(t)M_V(t)\\ &amp;=\left(M_U(t)\right)^2\\ &amp;=\left(e^{\mu t+\frac{1}{2}t^2\sigma ^2}\right)^2\\ &amp;=e^{2\mu t+t^2\sigma ^2}\\ \end{align*} The last expression is the moment generating function for a random variable distributed normal with mean $2\mu$ and variance $2\sigma ^2$.  Thus $U-V\sim N(2\mu,2\sigma ^2)$.  For the third line from the bottom, it follows from the fact that the moment generating functions are identical for $U$ and $V$.  Thanks for your input.  EDIT: OH I already see that I made a mistake, since the random variables are distributed STANDARD normal.  I will change my answer to say $U-V\sim N(0,2)$.
60615
I was reading  and I find the following problem: Let $X$ and $Y$ be independent $\chi ^2 $ with $m$ and $n$ degrees of freedom then $\dfrac{X}{X+Y} = \beta(m/2, n/2)$. Where $\chi^2(k)$ has density function $$f_{\chi^2(k)}(x)=\begin{cases}\displaystyle \frac{1}{2^{k/2}\Gamma(k/2)}\,x^{(k/2) - 1} e^{-x/2}&amp;\text{if }x&gt;0,\\ 0&amp;\text{if }x\le0 \end{cases} $$  I proved that $X+Y$ is $\chi ^2$ with $n + m$ degrees of freedom using the moment-generating function. If $Z=X+Y$ and $U=\dfrac{X}{X+Y}=\dfrac{X}{Z}$ using the formula for the density of the ratio of two independent random variables: $$f_U(u)= \int_{-\infty}^{\infty}f_X(uv)f_Z(v)|v|dv $$  then  $$\dfrac{u^{m/2-1}}{2^{(m+n/2)} \Gamma(m/2) \Gamma((m+n)/2)} \int_0^{\infty} v^{m+\dfrac{n}{2}-1} e^{-(v/2)(u+1)}dv $$  and using some algebra:  $$f_U(u)=\dfrac{u^{m/2-1} \Gamma(m+n/2)}{\Gamma(m+n/2) \Gamma(m/2) (u+1)^{m+n/2}}$$  and this is not $\beta(m/2, n/2)$, it looks more like $\beta(m/2, m/2+n/2)$ but $(u+1)^{m+n/2}$ is causing troubles.  I need help if something is false or any hint.
60870
Let $\ln X \sim \operatorname{Exp}(1)$ and $X,Y$ i.i.d.  What is the distribution of $Z=XY$?  My attempt: $$P(Z\le z)=P(XY\le z)=P(\ln(XY)\le \ln(z))=P(\ln(X)+\ln(Y)\le \ln(z))$$  We need the convolution, since $\ln X $ and $\ln Y$ are independent we get, $$\int_0^{\ln(z)} f_{\ln(X)}\circledast f_{\ln(Y)}=\int_0^{\ln(z)}e^{-y}e^{-x+y}\,dy=e^{-x}ln(z),$$ i.e. $e^{-x}\ln(z)$ is the density of $XY$
28457
I am trying to understand an example from my textbook.  Let's say $Z = X + Y$, where $X$ and $Y$ are independent uniform random variables with range $[0,1]$. Then the PDF is $$f(z) = \begin{cases} z &amp; \text{for $0 &lt; z &lt; 1$} \\ 2-z &amp; \text{for $1 \le z &lt; 2$} \\ 0 &amp; \text{otherwise.} \end{cases}$$  How was this PDF obtained?   Thanks
110892
Since integration is not my strong suit I need some feedback on this, please:  Let $Y$ be $\mathcal{N}(\mu,\sigma^2)$, the normal distrubution with parameters $\mu$ and $\sigma^2$. I know $\mu$ is the expectation value and $\sigma$ is the variance of $Y$.  I want to calculate the $n$-th central moments of $Y$.  The density function of $Y$ is $$f(x)=\frac{1}{\sigma\sqrt {2\pi}}e^{-\frac{1}{2}\left(\frac{y-\mu}{\sigma}\right)^2}$$  The $n$-th central moment of $Y$ is $$E[(Y-E(Y))^n]$$  The $n$-th moment of $Y$ is $$E(Y^n)=\psi^{(n)}(0)$$ where $\psi$ is the Moment-generating function $$\psi(t)=E(e^{tX})$$  So I started calculating:  $$\begin{align}  E[(Y-E(Y))^n]&amp;=\int_\mathbb{R}\left(f(x)-\int_\mathbb{R}f(x)dx\right)^n\,dx \\ &amp;=\int_\mathbb{R}\sum_{k=0}^n\left[\binom{n}{k}(f(x))^k\left(-\int_\mathbb{R}f(x)dx\right)^{n-k}\right]\,dx \\ &amp;=\sum_{k=0}^n\binom{n}{k}\left(\int_\mathbb{R}\left[(f(x))^k\left(-\int_\mathbb{R}f(x)dx\right)^{n-k}\right]\,dx\right) \\ &amp;=\sum_{k=0}^n\binom{n}{k}\left(\int_\mathbb{R}\left[(f(x))^k\left(-\mu\right)^{n-k}\right]\,dx\right) \\ &amp;=\sum_{k=0}^n\binom{n}{k}\left((-\mu)^{n-k}\int_\mathbb{R}(f(x))^k\,dx\right) \\ &amp;=\sum_{k=0}^n\binom{n}{k}\left((-\mu)^{n-k}E\left(Y^k\right)\right) \\ \end{align}$$  Am I on the right track or completely misguided? If I have made no mistakes so far, I would be glad to get some inspiration because I am stuck here. Thanks!
100312
Let $X_1,X_2,...,X_n$ be i.i.d with df $F(y)=y^{\theta}, 0&lt;y&lt;1, \theta&gt;0$. Show that $\frac{X_{(i)}}{X_{(n)}}$, for $i=1,2,...,n-1$ and $X_{(n)}$ are independent.  I found the population pdf to be $$f(x)=\theta x^{\theta -1} I_{0&lt;x&lt;1}$$ Now, $$f(x_1,x_2,...x_n)=\theta^n\prod_{i=1}^{n}{x_{i}}^{\theta - 1}$$ Now, the joint distribution of the order statistics: $$f_{1,2,..,n}(x_1,...x_n)=n! \theta^n\prod_{i=1}^{n}{x_{i}}^{\theta - 1} I_{0&lt;x_1&lt;x_2,...&lt;x_n&lt;1}$$ Now , we take the transformation $Y_i=\frac{X_i}{X_n} $ for $i=1,2,..,n-1$ and $Y_n=X_n$ Have I done correctly so far? Also, I found out the Jacobian of transformation $={y_n}^{n-1}$. I am also finding it difficult to get the ranges of $Y_i$'s. Help!
4380
I have to compute $Z(t)=\sum_{n=1}^\infty e^{-\lambda_nt}$, $t \in \mathbb{R}_{&gt;0}$, with $\lambda_n=\pi^2(\frac{m^2}{a^2}+\frac{n^2}{b^2})$. So $\sum_{m,n=1}^\infty e^{-(\pi^2(\frac{m^2}{a^2}+\frac{n^2}{b^2}))t}=\sum_{m=1}^\infty e^{-\pi^2(\frac{m^2}{a^2})t}\sum_{n=1}^\infty e^{-\pi^2(\frac{n^2}{a^2})t}$.  Is anyone could give me a good hint how to compute $\sum_{n=1}^\infty e^{-\pi^2(\frac{n^2}{a^2})t}$?
66693
I was reading this  where it had a gaussian distribution model.   I mean gaussian is given by  $P(y) = \frac{e^{-\frac{1}{2}(y -\mu)^T \Sigma^{-1}(y -\mu)}}{2\pi^{n/2}|\Sigma|^{1/2}}$  But is   $\frac{e^{-(y -\mu)^T \Sigma^{-1}(y -\mu)}}{2\pi^{n}|\Sigma|}$ Gaussian as well?  The paper used their model saying the gaussian had the above form. It is valid?
16379
Let $X$ and $Z$ be independent, with $X\sim N(0,1)$, and with $\textbf{P}(Z=1)=\textbf{P}(Z=-1)=\frac{1}{2}$. Let $Y=XZ$ (i.e., $Y$ is the product of $X$ and $Z$).   (a) Prove that $Y\sim N(0,1)$.  (b) Prove that $\textbf{P}(|X|=|Y|)=1$.   (c) Prove that $X$ and $Y$ are not independent.   (d) Prove that $\textbf{Cov}(X,Y)=0$.  (e) It is sometimes claimed that if $X$ and $Y$ are normally distributed random variables with $\textbf{Cov}(X,Y)=0$, then $X$ and $Y$ must be independent. Is that claim correct?   My Work: (a) Is really where I am a little stuck. In order to show that $Y\sim N(0,1)$, I want to show that they have the same distribution laws. Thus, I want to show that $\mathcal{L}(Y)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\textbf{1}_B(t)\lambda(dt)$ given that $\mathcal{L}(X)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\textbf{1}_B(t)\lambda(dt)$.   However, I don't know how to go about that. Another approach I have considered is to show that for any Borel measurable function $f:\mathbb{R}\to\mathbb{R}$,$\space$ $\textbf{E}(f(X))=\textbf{E}(f(Y))$ for which each is well defined. Which approach do you think is better and How should I start. Any help will be appreciated.
61409
Let $X,Y$ be two independent (and identically distributed) random variables. Let $Z:=X+Y$.  It's easy to check that the moment generating function $\phi_Z(t):=\mathbb{E}[\,e^{itZ}\,]$ can be expressed as $\phi_Z=\phi_X\cdot\phi_Y$.  Is there a way to express the cumulative distribution function $F_Z(z):=\mathbb{P}(Z\leq z)$ using the cumulative distribution functions of $X$ and $Y$ ?  Edit: note I don't assume that $X$ and $Y$ have a density.
52416
Suppose $X_1, X_2,$ and $X_3$ are independent and $N(0, 1)$-distributed. Compute the moment generating function of $Y = X_1X_2 + X_1X_3 + X_2X_3$.   I know that any $X_iX_j$ with $i \not =j $ is a joint normal with variables $(x_i,x_j)$ I also know the formula of the moment generating function of a normal distribution.  Furthermore, I know that if $Y_1,…, Y_n$ are independent $N(0,1)$, that is, $Y = (Y_1,…,Y_n )´$ are $N(0,I)$ by definition, the moment generating  function of Y is given by $$e^{\frac{1}{2}\mathbf t' \mathbf t}$$   I thought about using the pdf's and the definition of a moment generating function but it proved to be a really tedious process jacked of multiple integrations.  Does anyone know how to easily solve this problem with some relatively simple lines? (Especially using the multivariate normal properties and matrices)