Discrete random variable with absolute value and min

[dire remnant]

If X and Y follow a geometric distribution with parameter p and are independent.
Let S = min(X, Y) and T = |X - Y|
To calculate P(S = s ∩ T = 0)
Why is it not equivalent to calculating P(X + Y = 2s ) since min(X, Y) = 1/2(X + Y - |X - Y|)?
Thus, 2s = X + Y - 0

Thx for your help x)

[smoky junco]

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[turbid berry]

What are you stuck on?

[dire remnant]

why when i use my method i don't get this result
i have a factor (2s-1) and i don't understand why x')

[turbid berry]

Pr(X+Y=2s) includes the possibility that X=5 and Y=-5

However T would not be 0

[dire remnant]

oohhh oki thx ❤️
So we can't use the formula of the minimum for this kind of exercice ?

[grizzled blade]

You can prove that $(S=s) \cap (T=0)=(X=s) \cap (Y=s)$ both events are equal then use the fact that both random variables are independent

[dapper zinc]

😑 ɿotoЯ | Rotor 😑

[grizzled blade]

For the second one note that $(T=t)=(X-Y=t) \cup (Y-X=t)$

[dapper zinc]

😑 ɿotoЯ | Rotor 😑

[grizzled blade]

(The union is disjoint, try using the sigma additivity of the probability measure)

[dire remnant]

Oh, okay, thank you very much for your help 🙃 👍

[grizzled blade]

Np, for the second one, think a bit like the first one to what event is $(S=s)\cap (X=Y+t)$ equal to ? (same think for $(S=s)\cap(Y=X+t)$ )

[dapper zinc]

😑 ɿotoЯ | Rotor 😑

[grizzled blade]

If you think correctly in terms of indépendant random variables then you should be good

[grizzled blade]

same thing *

[dire remnant]

btw an another question
Do you think it's correct ?
(X and Y are two independent random variables with an exponential distribution with parameter lambda.)

[agile nova]

I think it'd be better to not use "probabilités totales" here if you don't have an explicit theorem for that, it's a bit wonky if you haven't seen this result in class

but that's just my personal opinion

$P(X \leq Y) = E[1_{{X \leq Y}}] = \int_{\mathbb{R}^2} 1_{{x \leq y}} f_{X,Y}(x,y)dxdy$

[dapper zinc]

Rion

[agile nova]

Then, you can decompose the joint density of (X, Y) using independence, and break the integral

to land back on what you got after that

[dire remnant]

oooh okay, thank you Rion 😁👍

[agile nova]

You are very welcome, please close this thread if you no longer need help

[dire remnant]

+close

[dusty slate]

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