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3

1) Draw the square $[-1,1]\times [-1,1]$. in the cartesian plane. The value of $X$ is on the $x$ axis, $Y$ on the $Y$ axis. 2) Asking the probability of $Z\leq s$ is equivalent to finding the area under the the line $x-y=s$, bounded by the square. 3) Find this area using simple geometry.


0

Expand the square. We get $\sum_1^n X_i^2 +2\sum_{1\le i\lt j}X_iX_j$. Now use the linearity of expectation, and independence. We get $nE(X_1^2)+(n)(n-1)E(X_1)E(X_1)$.


0

\begin{align*} &P(X+Y<Z)\\ &=\int_0^4 P(X+Y<z)\cdot \left( \frac{1}{4} \right)dz \\ &=\frac{1}{4} \left[\int_0^1 P(X+Y<z)dz+\int_1^2 P(X+Y<z)dz+\int_2^3 P(X+Y<z)dz+\int_3^4 P(X+Y<z)dz \right]\\ &=\frac{1}{4} \left[\int_0^1 \left(\frac{z^2}{2}\right) dz +\int_1^2 \left(z-1+\frac{1}{2}\right)dz +\int_2^3 ...


0

You tried to find $P(z<x+y)$ by $$ \int_0^1dx \int_0^1 dy \int_0^{x+y} 1\, dz $$ which does work out to 1. However, the range of $z <x+y$ is not $z\in(0,x+y)$ because $z$ cannot be more than 1. When $y>1-x$,$z$ can never exceed $x+y$, but you should integrate only up to $1$, not $x+y$. This explains the extra magnitude that brings that integral ...


1

We need the volume of $\{(X,Y,Z):\ Z<X+Y\}$ inside $[0,1]\times[0,2]\times[0,4]$. This is equal to $$\int_{0}^{2}\int_{0}^{1}(x+y)dxdy=\int_{0}^{2}\left(\frac{1}{2}+y\right)dy=3$$ Then divide by $1\times 2\times 4=8$ to get $$P(Z<X+Y)=\frac{3}{8}$$ The reason why this works is that the definition of uniform distribution in an interval is the usual ...


0

Do you know the correct answer? Because my answer to this problem is 0.5138170025.In other words, the probability of average waiting time to be less than 6 minutes is 51.38% Reply me whether I am correct or wrong.


0

The error covariance matrix is \begin{equation} \frac{1}{p+2} A^{-1}. \end{equation}


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I invite anyone who uses Mathematica to type in the following: F[n_] := (Times @@ RandomReal[{0, 1}, n])^(-n/2) Exp[n/2] Mean[Table[F[1000],{10^5}]] This gives the sample mean of $10^5$ simulations for $n = 1000$. It is a huge number. Now try F[10000], F[10^6], etc. The number of digits of these is enormous. Tell me how this is at all consistent with ...


1

Following snarski's hint, we have that $Y_i=-\log(X_i)$ has an exponential distribution with unit mean, hence: $$ \mathbb{P}\left[\log a \leq \frac{n}{2}\sum Y_i +\frac{n}{2}\leq \log b\right] $$ converges to zero since $\sum Y_i$ is concentrated around its mean value, $n$. Chebyshev's inequality is enough for proving it. Something significant is the ...


0

The density of $S$ is given by the convolution of the densities of $X$ and $Y$: $$f_S(s) = \int_{\mathbb R}f_X(s-y)f_Y(y)\mathsf dy. $$ Now $$ f_X(s-y) = \begin{cases} \frac12,& 0\leqslant s-y\leqslant2\\ 0,&\text{otherwise} \end{cases} $$ and $$ f_Y(y) = \begin{cases} \frac13,& 0\leqslant y\leqslant1\\ 0,&\text{otherwise.} \end{cases} $$ So ...


0

If $F$ denotes the CDF of $S$ then it is clear that $F(s)=1$ if $s\geq 5$ and $F(s)=0$ if $s\leq0$. Now let $s\in(0,5)$. Prescribe function $g_s$ on $\mathbb R$ by $\langle x,y\rangle\mapsto1$ if $x+y\leq s$ and $\langle x,y\rangle\mapsto0$ otherwise. $$F(s)=P(X+Y\leq s)=\int\int f_X(x)f_Y(y)g_s(x,y)dxdy=\frac16\int_0^2\int_0^3g_s(x,y)dydx$$ Here $f_X$ and ...


2

The area of the square is $X^2$. So you need to compute $\mathbb E[X^2]$ and $\mathrm{Var}(X^2)$. In general these quantities will not be equal to $\mathbb E[X]^2$ and $\mathrm{Var}(X)^2$. If $f$ is the probability density function of $X$ then $$\mathbb E[X^k] = \int_{-\infty}^\infty x^k f(x)\mathsf dx $$ for $k=1,2,3,\ldots$. Recall that ...


1

The distribution of extremal values is seldom normally distributed. The maximum of a bounded random variable will converge to a Type 3 Extreme Value Distribution (call ed weibull-law in the link). At the more basic level, the curvature of the log-likelihood at the extreme value is undefined, hence it will not converge to a quadratic log-likelihood.


2

Conditionally on $Y$, $X$ is uniformly distributed on the interval $[-|Y|,|Y|]$, and that distribution is symmetric about $0$, so $\operatorname{E}(X\mid Y)=0$ regardless of the value of $Y$. Conditionally on $Y$, $|X|$ is uniformly distributed on the interval $[0,|Y|]$. To see that, observe that for $0\le x\le |Y|$ we have $$ \Pr(|X|\le x\mid ...


-1

Hint: $$\operatorname{E}[|X|] = \operatorname{E}[\operatorname{E}[|X|\mid Y]] = 2 \int_{y=0}^a \operatorname{E}[|X| \mid Y = y] f_Y(y) \, dy.$$


1

Sorry, I didn't read your question carefully enough. The limit on the second of your integrals after doing the substitutions is wrong. Since $u=-x + (b+a)/2$ and $x$ runs from $(b+a)/2$ to $\infty$, you should have $u$ running from $0$ to $-\infty$. I think this is what has mislead you. The two integrals evaluate to $(3a+b)/8$ and $(a+3b)/8$, so their ...


1

Your first calculation is correct. Note that $$\sum_{i=1}^{2^n-1}P\left ( \left [ \frac{2i-1}{2^n}, \frac{2i}{2^n} \right ] \right )=\sum_{i=1}^{2^n-1} \frac{1}{\frac{2i}{2^n}-\frac{2i-1}{2^n}}$$ does not hold; that's where your second calculation fails. Remark: You can simplifiy the calculation by noting that $$A_n = \left[ \frac{1}{2^n}, 1 \right].$$



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