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0

There are two ways you could tackle the problem: (a) Use "change of variables" from $X$ and $V$ to $X$ and $Y$ and then find the required conditional distribution and expectation. (b) Use the "direct" method, which is what you've done. For this particular problem, I think (a) is a slightly simpler but I will use (b) here since it is in line with your ...


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If $U\sim U(0,1)$ then $f_U(u)=1 I_{[0,1]}(u)$ and $V\sim\Gamma(2,1)$ then $f_V(v)=ve^{-v}I_{[0,\infty]}(v)$ Now take $W=g_1(U,V)=UV$ and $Z=g_2(U,V)=V$ thus $$J=\begin{bmatrix}\frac{\partial W}{\partial U}&&\frac{\partial W}{\partial V}\\\frac{\partial Z}{\partial U}&&\frac{\partial W}{\partial V}\end{bmatrix}=v\Rightarrow ...


2

Brute force: let $S = UV$. Then $$\begin{align*} \Pr[S \le s] &= \Pr[UV \le s] = \Pr[V \le s/U] = \int_{u=0}^1 \Pr[V \le s/u] f_U(u) \, du \\ &= \int_{u=0}^1 \int_{v = 0}^{s/u} \frac{b^a v^{a-1} e^{-bv}}{\Gamma(a)} \, dv \, du \\ &= \frac{b^a}{\Gamma(a)} \left( \int_{v=s}^{\infty} \int_{u=0}^{s/v} v^{a-1} e^{-bv} \, du \, dv + \int_{v=0}^s ...


1

You have the right joint density function provided you get the right boundaries: $$ f_{S,T}(s,t) = \begin{cases} e^{-t} & \text{if }0<s<t, \\ 0 & \text{otherwise.} \end{cases} $$ (You can draw a picture and see that this is half of the first quadrant.) Now suppose we want the marginal distributions of $S$ and $T$. $$ f_S(s) = \int_s^\infty ...


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I am replying to Miller Zhu's comment (2nd comment to question), but the comment box doesn't have enough space. I'm only using the inscribed ellipsoid to get a sense of the principal axes of the polytope. Since you don't have access to the vertices of the polytope, computing the inscribed ellipsoid is the only computationally tractable problem (the ...


2

The easiest way: roll the dice twice. Now, there are 36 possible pairs. Delegate five pairs to each of 1, 2, 3, 4, 5, 6, 7 - then, if you get the remaining pair, roll the dice again. This guarantees equal probability for each of the 7 values. Note that the probability you don't have to roll again is $35/36$, so the expected value of dice you'll need to roll ...


1

You can't do it in a limited number of rolls, because in $N$ rolls, you get $6^N$ element in the sample space, and that isn't divisible by $7$. So roll the die twice, use those two values to encode a number from $1$ and $36$. If the first roll is $A$ and the second $B$, you can do this by computing $6A+B-6$. If you get $36$, roll again. If not, reduce ...


2

A simple counter-example can show dependence. For example: $P(0\lt u_1\lt \frac{1}{2} \;\cap\; 0\lt u_2\lt \frac{1}{2} \;\cap\; 0\lt u_3\lt \frac{1}{2}) = 0$ but $P(0\lt u_1\lt \frac{1}{2})P(0\lt u_2\lt \frac{1}{2})P(0\lt u_3\lt \frac{1}{2}) \gt 0$.


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We will need the joint comulatice distribution function $F_{Y_1,Y_2}(u,v)=P(Y_1<u,Y_2<v).$ Now, $$P(Y_1<u,Y_2<v)=P(Y_1<u,Y_2<v, X_1\le X_2)+P(Y_1<u,Y_2<v ,X_2<X_1)=$$ $$=P(X_1<u,X_2<v, X_1\le X_2)+P(X_2<u,X_1<v ,X_2<X_1).$$ First, $$P(X_1<u,X_2<v, X_1\le X_2)= \begin{cases} -\frac{u^2}{2}+uv,&\text{ if ...


1

My Trial $$\begin{align} F_Y(y) & = P(Y\leq y) \\[1ex] & = P(X^2\leq y) \\[1ex] & = P(X\leq\surd y)\\[1ex] & = F_X(\surd y) \\[1ex] & = \int_0^{\surd y} t \;\mathrm d t & \bigstar \\[1ex] & = 0.5 y\\[2ex] \therefore f_Y(y) & =0.5 \end{align}$$ $\bigstar$ Your error is here. $f_X(t) = 1$ for all $t\in[0;1]$, and $F_X(t) = ...


5

The method is fine in principle, but the wrong density function for $X$ was used..\ As in your argument, we have, for $0\lt y\lt 1$, $$F_Y(y)=\Pr(X\le \sqrt{Y}).$$ However, if you really want to use an integral, $$F_X(\sqrt{y})=\int_0^{\sqrt{y}} 1\cdot dt=\sqrt{y}.$$ This is because the density function of $X$ on $(0,1)$ is $1$. Differentiating, we find ...


2

You got it. Just follow through with that thought. $$\begin{split} P(\left| X-4 \right|) > 1.5) &= P(X > 5.5) + P(X < 2.5) \\ &= \frac{2.5}{6} + \frac{0.5}{6} \\ &= \frac12 \end{split}$$



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