Random point $(X, Y, Z)$ is uniformly distributed in the unit cub. What is the probability that $Y^2-4XZ \geq 0$?
12085
The following is a homework question for which I am asking guidance.     Let $A$, $B$, $C$ be independent random variables uniformly distributed between $(0,1)$. What is the probability that the polynomial $Ax^2 + Bx + C$ has real roots?   That means I need $P(B^2 -4AC \geq 0$). I've tried calling $X=B^2 -4AC$ and finding $1-F_X(0)$, where $F$ is the cumulative distribution function.  I have two problems with this approach. First, I'm having trouble determining the product of two uniform random variables. We haven't been taught anything like this in class, and couldn't find anything like it on Sheldon Ross' Introduction to Probability Models.  Second, this strategy just seems wrong, because it involves so many steps and subjects we haven't seen in class. Even if I calculate the product of $A$ and $C$, I'll still have to square $B$, multiply $AC$ by four and then subtract those results. It's too much for a homework question. I'm hoping there might be an easier way.
56745
A stick of unit length is cut into three pieces of lengths $X, Y, $and $Z$, with $X \le Y \le Z$. First the stick of length $1$ is cut into two pieces at a randomly chosen point uniformly distributed along its length. Then the smaller of these two sticks is also cut into two pieces at a randomly chosen point uniformly distributed along its length. These two cuts are conducted independently. Find Cov(X,Y).  I need to get $E(XY)-EXEY$. First, I write $S=X+Y$. Then the length function of $S$ is $\min(s, 1-s)$. Given $S=s$, I can find the covariance of $X,Y$ as follows. The distribution of $X$ is $\min(x, s-x)$, so $E[X|S]=\int_0^{s/2} xdx + \int_{s/2}^s (s-x) dx=s^2/4.$ Likewise, $E[Y|S]=\int_0^{s/2} (s-y)dy + \int_{s/2}^s y dy=3s^2/4.$ Now $E[XY|S]=\int_0^s x(s-x)dx=s^2/6$,since $y$ is just $s-x$(I'm not sure about this part). Now to get the expectation, I just need to integrate them over $s$, which has the uniform distribution, so that would be integrating them over $[0,1/2]$. Is this correct? I would greatly appreciate any comment.
4142
Given a random square matrix of size $n\times n$ in the field $\mathbb F_2$, what is the probability that its determinant is $1$? (This is also the probability that the matrix is non-singular, since $\mathbb F_2$ only has the elements $0$ and $1$.)
111736
It's known that .  Five random points can go right into the $6\times6$ matrix, and then the $A x^2 + B xy + C y^2$ part can be looked at.  If $B^2-4AC&lt;0$, it's an ellipse.  Five random points will almost never produce circles or parabolas, so the results will be ellipses and hyperbolas.  What are the odds of an ellipse?    In a random run of 100000 trials, I got 27974 ellipses. "It's less than $e/10$," seems like a solid answer.  Anyone have anything more specific?  EDIT:  As Oscar points out, I should have said "It's more than $e/10$."  In my trial, real-values points were randomly picked from a unit square.   methods might be applicable.  EDIT2: Aretino points out that  are $49/144≈0.34$. So how can points making a convex pentagon give a non-ellipse?  Here's a picture. With the red points fixed, the black points are outside of the convex hull yet still yield a non-ellipse.     EDIT3: That spray of points above goes back to Newton, Philosophiae naturalis principia mathematica, 1687, where he solved the  (). If a point is between one of the two parabolas and the degenerate lines, then it gives a hyperbola.
59841
If a point is randomly chosen from a square, whose sides are equal to 1, find the probability that the point's distance from a side of the square is less than the distance of that point from the closest square diagonal.   I am not sure how to picture this at all, really confusing. Any help is appreciated.
118167
I have the following problem:  Suppose that $Y \sim \text{Poi}(1)$. Let $y \ge 0$ be an integer. Given that $Y = y$, suppose that $X$ has the discrete uniform law on $\{ 0, 1, \dots, y \}$. (a) Show that $E[X \vert Y] = \dfrac{1}{2}Y$, and use this to find $E[X]$.  The provided solution is as follows:  We are given that $P(X = x \vert Y = y) = (y + 1)^{-1}$ for $x \in \{ 0, 1, \dots, y \}$, and $P(X = x \vert Y = y) = 0$ otherwise. (a) From the conditional PMF, we can compute the conditional expectation of $X$ given $Y = y$: $$\begin{align} E[X \vert Y = y] &amp;= \sum_{x = 0}^y xP(X = x \vert Y = y) \\ &amp;= (y + 1)^{-1} \sum_{x = 0}^y x \\ &amp;= (y + 1)^{-1} \left[ \dfrac{y(y + 1)}{2} \right] \\ &amp;= \dfrac{y}{2}. \end{align}$$ Therefore, $E[X \vert Y] = \dfrac{Y}{2}$. Now, since $Y \sim \text{Poi}(1)$, we know that $E[Y] = 1$. We now compute $E[X]$ using the law of total expectation: $$\begin{align} E[X] &amp;= E[E[X \vert Y]] \\ &amp;= E \left[ \dfrac{Y}{2} \right] \\ &amp;= \dfrac{1}{2} E[Y] \\ &amp;= \left( \dfrac{1}{2} \right) (1) \\ &amp;= \dfrac{1}{2}. \end{align}$$  How is it that $\sum\limits_{x = 0}^y x = \dfrac{y(y + 1)}{2}$? I would appreciate it if someone would please take the time to explain this.
8664
Let $X$ and $Y$ be independent random variables, uniformly distributed in the interval $[0,1]$. Find the CDF and the PDF of $|X - Y|$?  Attempt Let $Z = |X - Y|$, so for $z \geq 0$, the CDF $F_{Z}(z) = \mathbf{P}(Z \leq z) = \mathbf{P}(|X - Y| \leq z) = \mathbf{P}(-z \leq X - Y \leq z)$, which is where the algebra becomes confusing. Since they are independent, the joint pdf of $X$ &amp; $Y$ is simply 1, as long as $(X,Y)$ belong to the unit square. The solution suggests a plot the event of interest as a subset of the unit square and find its area. Any hints?
8299
I was reading Robert Serfling's 1980 book "Approximation Theorems of Mathematical Statistics" and came across the following construction of the Dvoretzky–Kiefer–Wolfowitz inequality for arbitrary distributions $F$,  which DKW prove for distributions on $[0,1]$.     Given independent $X_i$ with d.f. F and defined on a common probability space, one can construct independent uniform $[0,1]$ variates $Y_i$ such that $\mathbf{P}[X_i = F^{-1}(Y_i)] =1,\forall i$.   Why? Is this true for arbitrary distributions (including discontinuous ones)?  Secondly,     Let $G$ denote the uniform $[0,1]$ distribution and $G_n$ the sample distribution function of the $Y_i$s. Then $F(x)=G(F(x))$ and with probability $1$, $F_n(x) = G_n(F(x))$.   Why?  Now I don't understand quantile functions as well I would like, and so am having some trouble following these arguments.   Edit: All of this is on page 59 of the book.    whuber, thank you so much for your careful answer. It is appreciated. The answer to my question does indeed lie in the last paragraph of your reply - now if I could only wrap my head around it.  What is throwing me off is the following example which I have recreated from Galen Shorack's "Probability for Statisticians" (page 111). Here, the Lebesgue measure of the set $[X\neq F^{-1}Y]$ is not zero. Would you agree? I am referring to the points in the interval $(2,3)$ and in $(3,3.5)$ for which the inverse transformation does not bring any points back. Thank you again for looking into this.   \documentclass[a4paper]{article} % Graphics \usepackage{graphics} \usepackage{graphicx} \usepackage{pstricks} \usepackage{pst-plot} \usepackage{pstricks-add} \usepackage{epstopdf} \begin{document}  % An arbitrary CDF \begin{figure}[htbp] \begin{center} \begin{psgraph}[arrows=&lt;-&gt;](0,0)(-1.5,-.3)(6,1.2){.5\textwidth}{2.5cm}  \psplot[algebraic, linecolor=black]{-1}{1}{.025*x^2+.1*x+.175} % {this goes from .1 --&gt; .3}  \psplot[algebraic, linecolor=black]{1}{2}{-.1*x^2+.4*x+.1}  % {this goes form .4 --&gt; .5}  \psline[linecolor=black](2,.5)(3,.5)  % this stays at .5  \psline[linecolor=black](3,.6)(3.5,.6)  % this stays at .6  \psplot[algebraic, linecolor=black]{3.5}{5}{  -0.1333*(x-5)^2+.9}  % this goes from .6 --&gt; .9  \psdots[dotstyle=*](3,.6)(1,0.4)   \end{psgraph} \end{center} \caption{Arbitrary CDF with discontinuities and flat sections} \label{fig:cdf} \end{figure}  % The quantile function \begin{figure}[htbp] \begin{center} \begin{psgraph}[arrows=&lt;-&gt;](0,0)(-.3, -1.5)(1.2, 6){2cm}{5cm} % inverses using the Matlab finverse symbolic toolbox function \psplot[algebraic, linecolor=black]{.1}{.3}{20*(x/10 - 3/400)^(1/2) - 2}  \psline[linecolor=black](.3, 1)(.4, 1)   \psplot[algebraic, linecolor=black]{.4}{.5}{-((x-0.5)/(-0.1))^(0.5)+2}   \psline[linecolor=black](.5, 3)(.6, 3) \psplot[algebraic, linecolor=black]{.6}{.9}{-((x-.9)/(-0.1333))^(0.5)+5} \psdots[dotstyle=*](.6, 3)(.5, 2) \end{psgraph} \end{center} \caption{Quantile function of CDF in figure (\protect \ref{fig:cdf})} \label{fig:qf} \end{figure} \end{document}   PS. I could not use the comment box for the reply as I needed to use the &lt;code> environment.
80966
I define $X_i$ as a random variable that is uniformly distributed between (0,1). What is the expected number of such variables I require to make the sum go just higher than 1.  Thanks
90494
We consider a finite field $\mathbb{F}_q$ where $q=2p+1$ and $q$ and $p$ are prime numbers.  Let $r_i$ be a value picked uniformly at random from the field such that $r_i&gt;\frac{q}{2}$. Let $z$ be a fixed field element.   Question: Is $v=(\ (r_i)^{-1}\cdot z) \bmod q$, a uniformly random value of the field?
33721
Supppse $X$ and $Y$ are standard uniformly distributed in $[0, 1]$, and they are independent, what is the PDF of $Z = Y / X$?   The answer from some probability theory textbook is  $$ f_Z(z) =  \begin{cases} 1/2, &amp; \text{if } 0 \le z \le 1 \\ 1/(2z^2), &amp; \text{if } z &gt; 1 \\  0, &amp; \text{otherwise}. \end{cases} $$  I am wondering, by symmetry, shouldn't $f_Z(1/2) = f_Z(2)$? This is not the case according to the PDF above.
31967
What is the expected value of the maximum of 500 IID random variables   with uniform distribution between 0 and 1?   I'm not quite sure of the technique to go about solving something like this. Could anyone point me the right direction?  Thanks
55461
Let $X$, $Y$ and $Z$ be i.i.d. uniform (0,1) random variables. What is the distribution of $(XY)^Z$? I've tried to solve it via mgfs, and what I've found is:  $$E\left(e^{(XY)^Z}\right)=E\left(E\left(e^{(XY)^Z}|Z\right)\right)$$ $$=E\left(E\left(e^{(XY)}\right)^Z\right)$$ $$=E\left(2\left(\frac{e^t-1}{t}\right)^Z\right)$$ $$=2\int_0^1\left(\frac{e^t-1}{t}\right)^zdz$$ $$=2\int_0^1\exp\left({z\log\left(\frac{e^t-1}{t}\right)}\right)dz$$ $$=2\left[\frac{\exp\left({z\log\left(\frac{e^t-1}{t}\right)}\right)}{\log\left(\frac{e^t-1}{t}\right)}\right]_0^1$$ $$=2\frac{{\frac{e^t-1}{t}}-1}{\log\left(\frac{e^t-1}{t}\right)}$$  But I don't see where to go from here. It doesn't look like the mgf of any distribution that I know. How can I determine the distribution from this, or is there another way?
10126
I have to randomly generate 1000 points over a unit disk such that are uniformly distributed on this disk. Now, for that, I select a radius $r$ and angular orientation $\alpha$ such that the radius $r$ is a uniformly distributed variate from $r \in [0,1]$ while $\alpha$ is a uniformly distributed variate from $\alpha \in [0, 2\pi]$ using the following code r &lt;- runif(1000, min=0, max=1)  alpha &lt;- runif(1000, min=0, max=2*pi) x &lt;- r*cos(alpha) y &lt;- r*sin(alpha) plot(x,y, pch=19, col=rgb(0,0,0,0.05), asp=1)  Then I look at my sample space and it looks like this:  This obviously doesn't look like a sample with uniform distribution over the disk. Hence, I guessed that the problem might be occurring as a result of a lack of independence between the variables $r$ and $\alpha$ in contingency to how they've been linked computationally. To take care of that I wrote a new code. rm(list=ls()) r &lt;- runif(32, min=0, max=1) df_res &lt;- data.frame(matrix(c(-Inf, Inf), byrow = T, nrow = 1)) for (i in 1:32) {   for (j in 1:32) {     alpha &lt;- runif(32, min=0, max=2*pi)     r &lt;- runif(32, min=0, max=1)     df &lt;- data.frame(matrix(c(r[i],alpha[j]), byrow = T, nrow = 1))     df_res &lt;- rbind(df_res,df)   } } df_res &lt;- subset(df_res, df_res$X1 != -Inf) x&lt;- df_res$X1 *cos(df_res$X2) y &lt;- df_res$X1 *sin(df_res$X2) plot(x,y, pch=19, col=rgb(0,0,0,0.05), asp=1)  And, yet again the sample looks non-uniformly distributed over the disk  I'm starting to suspect that there is a deeper mathematical problem going on in the vicinity. Could someone help me write code that would create a sample space uniformly distributed over the disk, or explain the mathematical fallacy if any in my reasoning?
64136
Let $X,Y$ be independent $Uniform(0,1)$ random variables. Find $E(|X-Y|^a)$ where $a&gt;0$.  My working:  Define $W=1$ if $X&gt;Y$ and $W=0$ if $X&lt;Y$. We seek $E(|X-Y|^a)=E[E(|X-Y|^a|W)]=E(|X-Y|^a|W=1)P(W=1)+E(|X-Y|^a|W=0)P(W=0)=E((X-Y)^a)P(X&gt;Y)+E((Y-X)^a)P(X&lt;Y)$. Now $P(X&lt;Y)=\dfrac{1}{2}$ and $P(X&gt;Y)=\dfrac{1}{2}$ by symmetry. $E((X-Y)^a)=\int_{0}^{1}\int_{0}^{x}(x-y)^af_{X,Y}(x,y)dydx=\int_{0}^{1}\int_{0}^{x}(x-y)^adydx=\dfrac{1}{(a+1)(a+2)}$. Similarly $E((Y-X)^a)=\dfrac{1}{(a+1)(a+2)}$ by symmetry when $Y&gt;X$. Required expectation=$\dfrac{1}{(a+1)(a+2)}\dfrac{1}{2}+\dfrac{1}{(a+1)(a+2)}\dfrac{1}{2}=\dfrac{1}{(a+1)(a+2)}$  But correct answer is $\dfrac{2}{(a+1)(a+2)}$. Where am I going wrong?
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).
22029
$$P(X_1 &lt; X_2 &lt; X_3)= \frac{(1-p_1)(1-p_2)p_2p_3^2}{(1-p_2p_3)(1-p_1p_2p_3)}$$   To be frank I do not know where to start with this question, I would like an idea to get me going, or better yet an answer. This was on an earlier exam, I would like to be ready if something like this were to come up again tomorrow. What makes this a little bit more difficult than usual is the fact that geometric distribution is  a discrete variable..