# Geometric distribution

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Eris13696

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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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mqb2766

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#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)

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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Eris13696

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#3

(Original post by

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)

**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)

Last edited by Eris13696; 5 months ago

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mqb2766

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#4

(Original post by

I’m still confused, how are they equal to P(X=6) because when multiplied they don’t give that,

**Eris13696**)I’m still confused, how are they equal to P(X=6) because when multiplied they don’t give that,

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.

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Gregorius

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#5

**Eris13696**)

I’m still confused, how are they equal to P(X=6) because when multiplied they don’t give that,

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Eris13696

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#6

(Original post by

Don't multiply them, they're not independent! As mqb2766 points out, one implies the other.

**Gregorius**)Don't multiply them, they're not independent! As mqb2766 points out, one implies the other.

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mqb2766

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#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.

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

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#8

(Original post by

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.

**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.

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#9

(Original post by

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.

**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.

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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