$X\sim R(0,1)$ and $Y\sim R(0,1)$ (X and Y are uniformly distributed on the interval $[0,1])$  I need to find the density function of W when $W=X\cdot Y$   Can anyone help me?
68378
Say $X_1, X_2, \ldots, X_n$ are independent and identically distributed uniform random variables on the interval $(0,1)$.  What is the product distribution of two of such random variables, e.g., $Z_2 = X_1 \cdot X_2$?  What if there are 3; $Z_3 = X_1 \cdot X_2 \cdot X_3$?  What if there are $n$ of such uniform variables? $Z_n = X_1 \cdot X_2 \cdot \ldots \cdot X_n$?
118110
$X,Y\sim U(0,1)$  are independent, $W=\max (X,Y)$.      How do I find the PDF of $W$? How do I find the expectation of $W$ at two ways: 1. with the PDF of $W$ and without the PDF of $W$.   I ask this Q before few days, but now I want show you me solution and ask if it's right (and ask to tell me where I wrong):  We assume that $0&lt;t&lt;1$ becuase otherwise the function $(f)$ returns 0. 1. Lets say: $W=\max(X,Y)\Rightarrow W(t)=P(t&gt;X)\cdot P(t&gt;Y)=(1-P(t&lt;X))(1-P(t&lt;Y))$ because they are independent. We know that $P(t&lt;X)=F_X(t)=t$ (Because $X\sim U(0,1)$), and this is why: $F_W(t)=(1-t)^2$ (if $0&lt;t&lt;1$ otherwise is $0$). So $f_W(t)=F'_W(t)=2t-2$.  2: a. $$E(W(t))=\int _{0}^1 t\cdot f_W dt=\int_0^12t^2-2tdt=\left[\frac{2t^3}{3}-\frac{2t^2}2\right]^1 _0$$ I dot here something negative and I don't know why... Please tell me where my mistake here....  b. Here I'll show you the idea, and I'd like to know if I'm right...  $$\iint_{D_1}\max(X,Y)dxdy+\iint_{D_2}\max(X,Y)dxdy \\= \iint_{D_1}xdxdy+\iint_{D_2}ydxdy$$  I'm right? If I'm wrong please tell me where are my mistakes (at all the question). Thank you!
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.
62184
If $X=(x,y,z)$ is a random point uniform on the unit sphere in $\mathbb{R}^3$,  Are the  coordinates $x$, $y$, $z$ uniform in interval $(-1,1)$?
113413
Let $X$ be a random variable that follows $Uniform(-1,2)$ distribution.   Then what will be the cdf of $X^2$?   What i did was $$P(X^2\le y)=P(X\le y^{0.5})= \int_{-1}^{y^{0.5}}(1/3)dx=(1/3)(y^{0.5}+1) $$ $0\lt y \lt 4$ But this cdf doesnt give the probability $0$ at $y=0$ also the pdf which i get doesn't integrate to 1 What is wrong can somebody explain?? Am i doing something wrong??
51403
Let $X\sim N(0,1)$ and let $Z$ be a random variable independent of $X$ such that:  \begin{equation} \Pr(Z=z) = \begin{cases} \frac{1}{2} &amp; \mbox{if $z = -1$ or $z=1$}, \\ \\ 0 &amp; \mbox{otherwise}. \end{cases} \end{equation}   Define $Y = ZX$. Show that $Y \sim N(0,1)$. Determine $\operatorname{Cor}(X,Y)$. Are $X$ and $Y$ independent?   Progress:  I'm basically stuck at the beginning. I got as far as $P(ZX &lt; y) = P(Z &lt; y/X)$, which I'm not even sure is correct. I'm also not clear on what it means by defining $Y = ZX$ - does it mean I need to state the probability density function or something? And how do I show that $Y \sim N(0,1)$?
114727
Given $$f(x) = \begin{cases} x, \; 0 &lt; x &lt;1\\ 2-x, \; 1 \leq x &lt; 2\\ 0 \text{ everywhere else} \end{cases}$$  as our P.D.F, I must find the corresponding distribution function.   I know that $F(x) = P(X \leq x) = \int_{-\infty}^{x}f(t) dt$ is the distribution function, but I don't know how to apply it to this particular probability density function. The P.D.F has three different cases. How do I handle that? A sum of integrals with the appropriate bounds for each case of $f(x)$?  I'm kind of confused on how to create those bounds. How do we get rid the of the $-\infty$?  I'm thinking:  For $x &gt; 0$:  $$F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t)dt = \int_{0}^{1} x dx + \int_{1}^{2} 2-x dx$$ Eh....
70398
Let $X$ be a continuous random variable defined on the interval $[0,1]$ with density function $f_X(u)=c(1−u^2)$ for some suitable constant $c$.   Find: (a) the value of $c$, (b) (i) the expectation and (ii) the variance of $X$. Also find (c) the density function of the random variable $Y = X^{2}$.   Now I've already attempted (a) by integrating $(1-u^{2})$ between $0$ and $1$ (like $\pi$ in the Cauchy density function) and got $c=\frac{2}{3}$ and also for (b) (i) I integrated $uc(1-u^{2})$ between $0$ and $1$ which came out to be $c/4 = 1/6$.  Have I done these first $2$ parts correctly? If so how do I go about doing the rest and if not same question but also where did I go wrong?
119091
Let $X=\min(S,T)$ and $Y=\max(S,T)$ for independent exponential variables $S$ and $T$. Find the joint density of $X$ and $Y$. Are $X$ and $Y$ independent? How would you suggest I approach this?
4387
I tried finding the following density function:     Let $X$ and $Y$ be independent random variables, each having the exponential distribution with parameter $\lambda$. Find the joint density function of $X$ and $Z=X+Y$.   $\begin{aligned}f_{X,X+Y}=&amp;\frac{\partial}{\partial x\partial z}\mathbb P (X\leq x,Y\leq z-x)\\=&amp;\int_{-\infty}^x\int_{-\infty}^{z-x}f(u,v)\,\mathrm dv=\frac{\partial}{\partial z}\left[\frac{\partial}{\partial x}\int_0^x\int_0^{z-x}\lambda^2e^{-\lambda u-\lambda v}\,\mathrm dv\,\mathrm d u\right]\\ =&amp;\frac{\partial}{\partial z}\int_0^{z-x}\lambda^2e^{-\lambda x-\lambda v}\,\mathrm dv=\frac{\partial}{\partial z}\int_x^{z}\lambda^2e^{-\lambda x-\lambda(v-x)}\,\mathrm dv\\ =&amp;\lambda^2e^{-\lambda x-\lambda(z-x)}=\lambda^2e^{-\lambda z} \end{aligned}$  But I don't think this is correct, because $\mathbb P(X\leq x,X+Y\leq z)$ isn't equal to $\mathbb P(X\leq x,Y\leq z-x)$. But I can't think of anything else.  Could someone help me a little bit with this exercise?  I'm going to try the following:  $\mathbb P(X\leq x,Z\leq z)=\int_{-\infty}^x\int_{-\infty}^zf_X(u)f_Z(v)\,\mathrm du\,\mathrm dv$,  where I use the convolution formula. If I made no mistake, this gives  $f_{X+Y}(z)=z\lambda^2e^{-\lambda z}$, if $z&gt;0$.  Then calculating the joint density function:  $\begin{aligned}f_{X,X+Y}(x,z)=\frac{\partial}{\partial z}\left[\int_{-\infty}^z\lambda^3ve^{-\lambda x-\lambda v}\,\mathrm dv\right]. \end{aligned}$  Oh wait, this is also incorrect, because $X$ and $X+Y$ are dependent.  I really don't know how to proceed; could someone help?  I've also looked at this , but they mention the Jacobian, and my teacher has chosen to skip the paragraph about the Jacobian, so it wouldn't make sense if I had to use that.
109889
The continuous random variables $X$ and $Y$ have joint density function $f_{xy} = 1$ for $0 &lt; x &lt; 1$ and $2x &lt; y &lt; 2$, and zero otherwise.  I am stuck on the above question, with parts a and b below:  a) Find the $cov(x,y)$.  Work:   $cov(x,y) = E(XY) - E(X)E(Y)$   $= \int_0^1\int_{2x}^2 xydydx - \int_0^1\int_{2x}^2 xdydx\int_0^1\int_{2x}^2 ydydx$ = $\frac{-1}{18}$  However I am given the answer $\frac{1}{18}$.  Did I make a mistake in my computation or is the answer incorrect?  b) Find $E(X|Y = y)$.  Work:   $E(X|Y=y) = \int_{-\infty}^{\infty}xf_{x|y}(x|y)dx$  This is where I am puzzled.  I know that $f_{x|y}(x|y) = \frac{f_{xy}(xy)}{f_yy}$, and $f_yy= \int_{-\infty}^{\infty}f_{x,y}(x,y)dx$.  That means $f_yy = 1$ and therefore $f_{x|y}(x|y) = 1$.  However I am given the answer $E(X|Y=y) = \frac{y}4$. Again, did I make a computational mistake or am I not getting something?  Thanks.
16940
Suppose that the continuous random variable $X$ has probability density function  $f_X(x)=\begin{cases}\frac{1}{2}x &amp; \text{if } 0&lt;x&lt;2\\0&amp;\text{otherwise}\end{cases}$  Let $Y=X(2-X)$. Calculate $P(Y&gt;y)$ and hence find the probability density function of $Y$.  So far I have tried expressing $P(Y&gt;y)$ as $P(X(2-X)&gt;y)$ but the quadratic inside has me stuck. Sorry - I am revising for an exam and it has been a long time since I have answered questions like this.
31303
I am calculating the sum of two uniform random variables $X$ and $Y$, so that the sum is $X+Y = Z$. Since the two are independent, their densities are $f_X(x)=f_Y(x)=1$ if $0\leq x\leq1$ and $0$ otherwise. The density of the sum becomes $f_Z(z)=\int_{-\infty}^\infty f_X(z-y)f_Y(y)dy=\int_0^1f_X(z-y)dy$ by convolution. I am stuck at this stage. How do I proceed with my integral? I think a diagram make it easy but I dont know how to proceed.
110924
I was wondering if someone could explain the following passage from my textbook.    In the third paragraph from the bottom it is stated that the simulated values come from the marginal distributions. How is this so? We simulate the value x2 from the condition density f(x2|x1)..  Also, there is mention of independence and again because of the dependence of the value of x2 on x1 and so forth, I do not see how this occurs.   Would greatly appreciate your help
115471
Intuitively, I understand that if $Y$ is a constant random variable and $X$ is another random variable, then $X$ and $Y$ are independent.   However, I can't make a formal proof because I can't show that their joint density function are the product of two functions that rely only on x's and y's respectively or using similar methods.   (What is the density function of a constant random variable for example?)  Can you give me a hint in order to make a proof?
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.
53507
If $\mathbf{x}$ and $\mathbf{y}$ are two independent random unit vectors in $\mathbb{R}^D$ (uniformly distributed on a unit sphere), what is the distribution of their scalar product (dot product) $\mathbf x \cdot \mathbf y$? I guess as $D$ grows the distribution quickly (?) becomes normal with zero mean and variance decreasing in higher dimensions $$\lim_{D\to\infty}\sigma^2(D) \to 0,$$ but is there an explicit formula for $\sigma^2(D)$? Update I ran some quick simulations. First, generating 10000 pairs of random unit vectors for $D=1000$ it is easy to see that the distribution of their dot products is perfectly Gaussian (in fact it is quite Gaussian already for $D=100$), see the subplot on the left. Second, for each $D$ ranging from 1 to 10000 (with increasing steps) I generated 1000 pairs and computed the variance. Log-log plot is shown on the right, and it is clear that the formula is very well approximated by $1/D$. Note that for $D=1$ and $D=2$ this formula even gives exact results (but I am not sure what happens later).