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2

Based on your second paragraph, I assume you ask if the following is true or not: $$p(x_1|x_2,x_3)p(x_2|x_3)p(x_3)=p(x_3|x_2,x_1)p(x_2|x_1)p(x_1)$$ Yes, it holds. The indices doesn't matter. Call them $x_1=a,x_2=b,x_3=c$ and you won't have a standard form. To be specific, you'll have $n!$ different factorizations if you have $n$ RVs. The intuition is you ...


2

In literature, normalization means integrating to $1$, not having a max value equal to $1$. So, joint or univariate densities are already normalized. For the nomenclature, for the function you have, I think max-normalized joint density would be a better name for it. However, what you do is just scaling your joint PDF so that it hits $1$ at its maximum. ...


2

No, and it is quite easy to see why. If you think of plotting the marginals on two separate axis in 2-D space to obtain the joint, you can easily define any esoteric function which maintains their marginal distributions when brought back to the axis, yet has an arbitrarily defined joint distribution. And even with a defined covariance matrix it is not enough ...


5

The support is the rectangle $\mathcal{R}=[0,3]\times[0,2]$, and for $P(X>Y)$, you'll integrate the area under $Y=X$ line: $$\int_{\mathcal R} f_{X,Y}(x,y)dydx=\int_{0}^1\int_0^x \frac{1}{9} dy dx+\int_1^2\int_0^x\frac{2}{9} dydx$$


2

$P(A, B)$ is a pretty standard notation for joint probability two events $A$, $B$, both being true, $P(A \cap B)$ means the same, but it is using set theory notation for intersection, in probability theory it is just as a notation used for saying $A$ and $B$, $P(A \land B)$ is the same, but is using a notation borrowed from logic for conjunction, $P(AB)$ or $...


3

I think you used the Cholesky decomposition where $L$ is a lower triangular matrix, meaning $b=0$. I hope this helps.


0

$P(AB)$ is not a standard notation, $AB$ can be the name of the event, or as in your example, each letter can denote different events. But, $P(A,B)$, which maybe used for shorthand purposes, and others most probably mean the same thing.


1

The converse is also true and you can write the joint pmf as the multiplication of the two marginal pmfs.


3

By definition, it's Chi squared with two degree of freedoms. The one you've found is the joint PDF of $U$ and $V$. You need to marginalize it to obtain $f_U(u)$: $$f_U(u)=\int_{-\sqrt u}^\sqrt u \frac{1}{\sqrt{u-v^2}}\frac{e^{-u/2}}{2\pi}dv=\frac{e^{-u/2}}{2\pi}\overbrace{\int_{-\sqrt{u}}^{\sqrt u}\frac{1}{u-v^2}dv}^\pi=\frac{1}{2}e^{-u/2}$$ where $u\geq 0$. ...


2

If I'm not mistaken, this should have chi-square distribution with 2 degrees of freedom. If X and Y are standard normal, then this should follow from the definition of the chi square distribution. See here.


0

You can write what you ask (instead of $<$, you'll need $\leq$ whenever CDF is concerned, especially with discrete RVs) in terms of CDFs: $$P(X>6,Y\leq7)=P(Y\leq 7)-P(X\leq 6, Y\leq 7)=F_Y(7)-F_{XY}(6,7)$$ And, your formula is correct only when $X\leq x_2$ and $Y\leq y_2$. And, you can calculate the marginal CDF of $Y$, using the joint CDF and the ...


0

This is due to the definition of distance in the sense of metric space: In mathematics, in particular geometry, a distance function on a given set $M$ is a function $d: M \times M \to \mathbb{R}$, where $\mathbb{R}$ denotes the set of real numbers, that satisfies the following conditions: $d(x,y) \ge 0$, and $d(x,y) = 0$ if and only if $x = y$. (...


2

For an affine transformation of a random vectors, the following rules apply. Let $\mathbf{X}$ be the $n\times 1$ column vector with random variables $X_1, X_2, \ldots, X_n$ as its elements. Let $\mathbf{a}$ and $\mathbf{b}^T$ be given matrices of size $k\times 1$ and $k\times n$ then $$ \mathbf{Y} = \mathbf{a} + \mathbf{b}^T \mathbf{X} $$ is a random ...


0

The last variable is linearly dependent on the first $n$. Let the joint PDF of the original random vector be $f_{\mathbf{X}}(\mathbf{x})$. And, denote the new PDF as $f_{\mathbf{X}_y}(\mathbf{x},y)$. Then, since the linear relation between $x_i$ and $y$ must hold: $$f_{\mathbf{X}_y}(\mathbf{x},y)=f_\mathbf{X}(\mathbf{x})\delta\left(y-\sum_{i=1}^n x_i\right)$$...


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