What is the average distance of two points chosen uniformly on a unit square? What I am asking is how to calculate $E\left(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\right)$ for $x_1, x_2, y_1, y_2$ spread uniformly on $[0,1]$.
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A square with side $a$ is given. What is the average distance between two uniformly-distributed random points inside the square?  For more general "rectangle" case, see . The proof found there is fairly complex, and I am looking for a simpler proof for this special case. I expect it could be significantly simpler.  See also .
120056
[] ** calculate the mean and standard deviation of each investment, assume that both returns come from independent samples, are both normally distributed and that the (true but unknown) variances are the same,, Can someone could check my  calculations for  mean and standard deviation? I did calculations but not sure what to do with negative values of distances(If if am correct Investment A MEAN= 78.6875 when it comes to calculating standard deviation i have to power difference by 2 ?  **
112318
what is wrong with this??? please,can anyone help?  \begin{equation} \label{probabilityy} P^{t}({E_{0})=P_{0}+(1-P_{0})P_{1}+ (1-P_{0})(1-P_{1})P_{2}+.......+P_{n}\Pi_{k_{0}}^{n-1}(1-P_{k}) \end{equation}
84058
If the best linear approximation (using least squares) of my data points is the line $y=mx+b$, how can I calculate the approximation error? If I compute standard deviation of differences between observations and predictions $e_i=real(x_i)-(mx_i+b)$, can I later say that a real (but not observed) value $y_r=real(x_0)$ belongs to the interval $[y_p-\sigma, y_p+\sigma]$ ($y_p=mx_0+b$) with probability ~68%, assuming normal distribution?  To clarify:  I made observations regarding a function $f(x)$ by evaluating it a some points $x_i$. I fit these observations to a line $l(x)=mx+b$. For $x_0$ that I did not observe, I'd like to know how big can $f(x_0)-l(x_0)$  be. Using the method above, is it correct to say that $f(x_0) \in [l(x_0)-\sigma, l(x_0)+\sigma]$ with prob. ~68%?
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Two numbers are chosen randomly and independently from the unit interval $(0,1)$.  What is the probability that they differ by more than $\frac{1}{2}$? Hint: Think of the $2$ numbers chosen as coordinates in the Cartesian plane. Draw a square with vertices at $(0,0)$,$(1,0)$,$(1,1)$, and $(0,1)$. Allowable choices for the $2$ points lie below the line $y=x-\frac{1}{2}$ or the above $y=x+\frac{1}{2}$. Find the combined area of these regions.  I need help on this. I know what $y=x-\frac{1}{2}$ and $y=x+\frac{1}{2}$ looks like but I am not understanding the problem as a whole
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$X,Y$ and $Z$ are independent uniformly distributed on $[0,1]$  How is random variable $(XY)^Z$ distributed?  I had an idea to logarithm this and use convolution integral for the sum, but I'm not sure it's possible.
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Suppose I have $N$ points on a unit circle. I am aware that for a uniform distribution the average angle of one point is not defined, but I want to calculate the average arc length between nearest-neighbors. I disregard arc lengths larger than $\pi$.  For $N = 2$ I am formulating the problem as: \begin{equation} \langle \Delta x \rangle = \frac 1 {\pi^2}\int\limits_0^\pi \int\limits_0^\pi  dx_1 dx_2 \, |x_1-x_2| = \frac \pi 3. \end{equation}  This answer does not make sense to me. Did I make a mistake constructing this integral?  What would happen to $N = 3$, etc?
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I came across the following problem in my self-study of real analysis:     For any real numbers $a$ and $b$, show that $$\max \{a,b \} = \frac{1}{2}(a+b+|a-b|)$$ and $$\min\{a,b \} = \frac{1}{2}(a+b-|a-b|)$$   So $a \geq b$ iff $a-b \ge0$ and $b \ge a$ iff $b-a \ge 0$. At first glance, it seems like an average of distances. For the first case, go to the point $a+b$, add $|a-b|$ and divide by $2$. Similarly with the second case.   Would you just break it up in cases and verify the formulas? Or do you actually need to come up with the formulas?
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If $X_1,\ldots,X_n\sim Uniform(0,1)$ and $Y=\max[X_1,\ldots,X_n]$, find $E(Y)$.   I don't really have any good starting point. I guess Y is a function of the X's and so I might be able to apply the theorem about the expectation of a function, but I don't know how to deal with this since max is a discrete function.  Can anybody help me please?
110481
How can we calculate the distance from a point to a plane in normed spaces ? where we are not inner product, for example:  Calculate the distance in $(C[0,1],||\cdot||)$ endowed with the supremum norm $$||f||=\sup_{t\in[0,1]}|f(t)|$$ from the function $f(t)=t^2$ to the linear hull of the functions $g(t)=t$  and  $h(t)=\sin(t)$  Any hints would be appreciated.
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$X\sim U[0,1]$ and $Y\sim U[-1,1]$ are two uniform-distributed R.V.'s. Are $X^2$ and $Y^2$ still uniform? Do they have explicit probability density funtion?
8153
I have 2 random variables $X,Y$ with mean 0 and variance 1, their correlation is $\rho$. I need to prove this inequality  $$\mathrm E\left({\max}\left\{X^2 , Y^2\right\} \right) \leq 1 + \sqrt{1 - \rho^2}$$ I need some pointers as to how to solve this problem. Thanks!
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I'm looking for a simple argument to the following:  Given two (continuous) paths on the unit square, one from (0,0) to (1,1) and the other from (1,0) to (0,1), prove that the paths cross at some point $(x_0, y_0)$.  I have constructed a topological argument for why this is true using compactness (and a proof by contradiction), but the person I'm working with seems to think there is a "simple" and "well-known" argument that says the two paths must cross.  I haven't been able to find such a result.  Does anyone know of one?  Thanks!
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?
82578
What is the Orthogonal Projection onto the $ {\ell}_{\infty} $ Unit Ball?  Namely, given $ x \in {\mathbb{R}}^{n} $ what would be:  $$ {\mathcal{P}}_{ { \left\| \cdot \right\| }_{\infty} \leq 1 } \left( x \right) = \arg \min_{{ \left\| y \right\| }_{\infty} \leq 1} \left\{ {\left\| y - x \right\|}_{2}^{2} \right\} $$  I managed to get an answer using the . Yet I would be happy to see if someone can derive the answer directly.  Thank You.
53462
What is the variance of the sample variance? In other words I am looking for $\mathrm{Var}(S^2)$.  I have started by expanding out $\mathrm{Var}(S^2)$ into $E(S^4) - [E(S^2)]^2$  I know that $[E(S^2)]^2$ is $\sigma$ to the power of 4. And that is as far as I got.
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Basic Question  Is there an intuitive explanation of standard deviation in terms of Euclidean distance in $n$ dimensional space?  Longer Version of Question  To begin a more detailed sketch of my question, for simplicity let's just focus on the simple case of a discrete random variable that is uniformly distributed. In this case, the variance is given by the following formula, which I've abducted straight from Wikipedia:  $$ \frac1{n}\sum_{i =1}^n (x_i - \mu)^2$$  where $\mu$ is the mean. The standard deviation is then the square root of this. Now, I can't help noticing that the square root of the sum returns the euclidean distance from the vector $X = (x_1, x_2, \dots, x_n)$ to the vector $\vec \mu = (\mu, \mu, \dots, \mu)$. That is, the standard deviation can be expressed as:  $$ \frac1{\sqrt{n}}|X - \vec \mu |$$  So I wonder, is there any significant conceptual relationship between this distance $|X - \vec \mu |$ and standard devation or is this just a coincidence?  Even More Details...  I have looked up many explanations of standard deviation and its cousin variance. Here are some that I've seen already, each sort of following from the previous one:   We square the values before summing to get rid of the sign, which is obviously not important. This explanation is often criticised by hardcore statisticians and I can sort of see why: it doesn't explain why squaring beats taking the absolute value. We square the values so that we pay a greater price for greater deviations. This explains why squaring beats taking absolute values. But why not raise to the power of $4$, or $6$, or any other even power before summing? What is so special about $2$? The thing that is so special about $2$ is that it's the second moment of intertia, whereas the mean is the first moment, so mechanically it makes sense. I don't follow this. My intuition is totally OK with the mean: the point where, if I put my finger, the weights on either side will balance. But the second moment is harder for me to imagine physically like this.   Note, this is a question about intuition. I "understand" the mathematical formula at a shallow level: what all its terms mean, how to calculate it given a dataset. But I am not comfortable with my grasp on why this formula is "the best" one to use in so many applications e.g. the least squares method to fit data. I'm particularly confused as to why squaring is chosen as opposed to raising to some other even power e.g. $9234324$.   And this is where my intuition steps in and tries to provide an explanation that goes right back to the fundamental theorem of Pythagoras: euclidean distance. Here is my thought process: "The number $2$ is special. It's the unique power that makes Euclidean distance work. So maybe it's also the unique number that makes variance work." But then why the multiplying factor of $\frac1{\sqrt{n}}$? Is it just simply a case of: swallow it up and accept the definition, or can this intuition be resolved somehow?