Variance distrubution -Geometric vs normal

If var(x)=2
Normally var(5x) would be the same as 25var(x) so 25x2=50
But for a geometric distrubution this would be 5var(X) so 5x2=10 is these a reason for that because I,m very confused

Reply 1

DFranklin

Why do you think that "for a geometric distrubution this would be 5var(X)"?

Reply 2

RDKGames

Study Forum Helper

Original post by realed

If var(x)=2
Normally var(5x) would be the same as 25var(x) so 25x2=50
But for a geometric distrubution this would be 5var(X) so 5x2=10 is these a reason for that because I,m very confused

No wonder you’re confused, that’s incorrect.

Reply 3

realed

Original post by RDKGames

No wonder you’re confused, that’s incorrect.

I’m the mark scheme it said it else’s that

Reply 4

mqb2766

Original post by realed

I’m the mark scheme it said it else’s that

Probably help to post both the question and the mark scheme.

Reply 5

DFranklin

In the absence of the question or mark scheme, I'll make the following guess at the issue:

I suspect that you've assumed taking the sum of 5 independent identically distributed (iid) random variables is the same as taking one variable and multiplying by 5; this is not the case.

[If

X_1, X_2, X_3, ..., X_5

are independent random variables, then

Var[X_1+X_2+...+X_5] = 5Var[X_1]

, while in contrast

Unparseable latex formula:

Var[5X_1] = 25 \Var[X_1]

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Variance distrubution -Geometric vs normal

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