# Geometric distribution

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#1
What do they mean in d) “if x=6, x>4 adds nothing”, why does it add nothing?
Last edited by Eris13696; 5 months ago
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5 months ago
#2
The probability of being equal to 6 and > 4 is the same as P(X=6). The exta condition > 4 does not change it.

As an altenative question related to this, work out
P(X>4 | X=6)
Last edited by mqb2766; 5 months ago
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#3
(Original post by mqb2766)
The probability of being equal to 6 and > 4 is the same as P(X=6). The exta condition > 4 does not change it.

As an altenative question related to this, work out
P(X>4 | X=6)
I’m still confused, how are they equal to P(X=6) because when multiplied they don’t give that,
Last edited by Eris13696; 5 months ago
0
5 months ago
#4
(Original post by Eris13696)
I’m still confused, how are they equal to P(X=6) because when multiplied they don’t give that,
Yes
P(X=6 and X>4) = P(X=6)
That is what the question/answer says in the book.

For the "example" I gave, it should be clear (by definition and cond0tional probability equation) that
P(X>4 | X=6) = 1
Given that X=6, then it is > 4 with absolute certainty.
1
5 months ago
#5
(Original post by Eris13696)
I’m still confused, how are they equal to P(X=6) because when multiplied they don’t give that,
Don't multiply them, they're not independent! As mqb2766 points out, one implies the other.
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#6
(Original post by Gregorius)
Don't multiply them, they're not independent! As mqb2766 points out, one implies the other.
Oo I see, I thought the geometric distribution probabilities are independent, does that only apply to the probability of p and q and not the probabilities we are finding?
0
5 months ago
#7
I guess you're talking about the independent trials used in the definition of the basic geometric distribution? Here the joint distribution is different and not independent, in fact its a variant of something like
P(X=6 and X=6) = P(X=6)
P(X=6 | X=6) = 1

If you assume independence,
P(X=6 and X=6) = P(X=6)^2
which can't be true in general as it must be P(X=6).

The independence (or not) of a joint distribution can't really be inferred from the marginal (single variable) distributions.
Last edited by mqb2766; 5 months ago
0
#8
(Original post by mqb2766)
I guess you're talking about the independent trials used in the definition of the basic geometric distribution? Here the joint distribution is different and not independent, in fact its a variant of something like
P(X=6 and X=6) = P(X=6)
P(X=6 | X=6) = 1

If you assume independence,
P(X=6 and X=6) = P(X=6)^2
which can't be true in general as it must be P(X=6).

The independence (or not) of a joint distribution can't really be inferred from the marginal (single variable) distributions.
Ooh that makes sense, does this apply to all questions that come this way? Because for now I haven’t really been introduced to joint distribution so that’s why I’m a bit confused on the idea of what this is.
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5 months ago
#9
(Original post by Eris13696)
Ooh that makes sense, does this apply to all questions that come this way? Because for now I haven’t really been introduced to joint distribution so that’s why I’m a bit confused on the idea of what this is.
If you've not really come across independence / joint distributions, then I'd not worry about it too much and come back to it when you do it properly.

For this one, you (can) get the joint using the conditional
P(a and b) = P(a | b) P(b)
and if they were independent you'd have P(a | b) = P(a). In this case you have P(a | b) = 1, so they're not independent.
Last edited by mqb2766; 5 months ago
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