New answers tagged uniform-distribution
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We need to assume something about the relationship between $X$ and $Y$. For example, if $X=Y$, then $X-Y$ is identically $0$. We will assume what we are expected to assume but should have been mentioned, that $X$ and $Y$ are independent.
We will first find the cumulative distribution function $F(z)$ of $Z$, that is, the probability that $Z\le z$. First ...
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Here is how I would start:
$$P(Z \le z) = P(X-Y \le z) = \int_0^a P(X \le z+y) f(y) dy$$
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For each point, independently generate two numbers: $l_i\sim U(0,m)$ and $h_i\sim U(0,n)$. The resulting point $(l_i,h_i)$ is a point in the rectangle and the joint distribution is uniform over the entire rectangle.
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The proof asserts: $P(2X\gt Z) = \min\{{2X,1\}}$
and the OP asks why this is so. Neither the proof nor the answer above make any sense to me.
If X ~ Uniform(0,1) and Z ~ Uniform(0,1) are independent, then $P(2X\gt Z)$ = $\frac{3}{4}$.
By contrast, $\min\{{2X,1\}}$ = 2X if x < $\frac{1}{2}$, and 1 otherwise. It is not a constant, nor a probability: it ...
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If $2X>1$ then $P(2X>Z)=1$. If $2X<1$ then $P(2X>Z)=2X$, since $Z$ is uniform on $[0,1]$.
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There are different solutions depending on whether a > 0 or a < 0, c > 0 or c < 0 etc. If you can tie it down a bit more, I'd be happy to compute a special case for you.
In the case of: a > 0 and c > 0, there are three special sub-cases:
Case 1: a d > b c
Case 2: a d < b c
Case 3: a d = b c
Here is the output for case 1 using the ...
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The general formula for the pdf of the product $Z=XY$ of two random variables $X$ and $Y$ is given by
$$p_{Z}(z)= \int_{-\infty}^{\infty}\frac{1}{|x|}p_{XY}\left (x,\frac{z}{x}\right )\;dx$$
where $p_{XY}(x,y)$ is the joint distribution of the two random variables. If they are independent you can use $p_{XY}(x,y)=p_X(x)p_Y(y)$.
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