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1 vote
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Finding CDF of function of random variables

Your computations are almost correct, there is only a subtle error. In order to have $XY\leq e^t$, given that $X\geq 1$ we must have $Y\leq e^t$. Therefore, the correct integral should be $$F_Z(t)= \...
Julio Puerta's user avatar
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1 vote

Constructing a density function from a condtion

Write your functional equation as $$ f(x) = \frac{1/\nu - \theta h f(xh)}{1-\theta}$$ You have $f(x) = 0$ for $x > \nu$. Thus for $\nu \ge x > \nu/h$ you must have $f(x) = 1/(\nu (1-\theta))$. ...
Robert Israel's user avatar
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Related to CDF of product of two independent Gamma random variables

It's quite hard to get an explicit formula for CDF of a product of Gamma r.v.s. One of the ways is to use the Mellin transform, which has a great condition, that if you have 2 random variables, then ...
BigFun's user avatar
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0 votes
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What does the values of Joint Probabilities indicate here?

It simply means $$f(x,y) = \begin{cases} \frac{1}{2}, & 0 \le y \le x \le 1 \\ \frac{3}{2}, & 1 < x \le x+y \le 2 \\ 0, & \text{otherwise}. \end{cases}$$ It's not actually clear what ...
heropup's user avatar
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0 votes
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Solving a stochastic equation by characteristic functions

I got it. We have to use the Mellin-transform mentioned in the Wikipedia article about the distribution of the product of two random variables. The Wikipedia-article quotes the property, that for ...
user1047209's user avatar
0 votes

Calculate probability density function of a random vector $(X + Z, Y + W)$

The form of the density implies that the conditional distribution of $Y$ given $X$ is that of a normal distribution with mean $X$ and variance $X$, and that the marginal distribution of $X$ is that of ...
Julius's user avatar
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Density function of the sum of 2 random variables

In case you are not up on Jacobian transformations, you could consider the joint cdf for $X+Y$ and $Y$. $$ F_{X+Y,Y}(u,v) = P(X+Y \leq u, Y \leq v) $$ You can compute this by integrating over the ...
aprobabilityspace's user avatar
2 votes

Switching limits of a pdf

The region is a parallelogram. You only need one integral if you integrate with respect to $y$ first. For each $x$, $y$ goes from $x$ to $x+1$ ,so the integral should be $$ \int_{0}^{1} \int_{x}^{x+1} ...
aprobabilityspace's user avatar
3 votes
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Switching limits of a pdf

For a question like this, it's almost always a good idea to sketch the region in the $xy$ plane. Now if you look at how lines of constant $y$ cut this parallelogram, you see that you need different ...
Robert Israel's user avatar
1 vote

Phase distribution of a non-zero mean complex Gaussian random variable

One can also find a derivation of this probability density function in Appendix D of https://arxiv.org/abs/2402.05793v1 Here we provide a complete derivation of the Rician phase probability density ...
Mark M. Wilde's user avatar

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