X has the Cumulative distribution function:
0 for x<0
x/4 for 0≤x≤1
3x/4  1/2 for 1<x≤2
1 for x>2
But how do I know which one to choose in order to get the medianF(m)=0.5?
Apparently it's the '3x/4  1/2 for 1<x≤2' one.

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 1
 22032013 15:05

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 22032013 15:42
(Original post by zomgleh)
X has the Cumulative distribution function:
0 for x<0
x/4 for 0≤x≤1
3x/4  1/2 for 1<x≤2
1 for x>2
But how do I know which one to choose in order to get the medianF(m)=0.5?
Apparently it's the '3x/4  1/2 for 1<x≤2' one.
If F(1) is < 0.5 then you know it's in the second interval.
If it's > 0.5 it would be in the first interval. 
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 22032013 16:46
Cheers.
Got another one:
if the CDF is the following:
F(x) = 0 when x<0
1(2/3)e^(0.5x) when 0≤x≤2
1 when x>/= 2
Then this represents a mixed distribution both discrete and continuous, so if we were to calculate P(X=2), the answer reads 2/3 e^1 which probably means the 2nd interval is being used with 1F(2). I was wondering why this is the case? Shouldn't the P(X=2) be F(2)F(0)? 
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 4
 22032013 17:10
(Original post by zomgleh)
Cheers.
Got another one:
if the CDF is the following:
F(x) = 0 when x<0
1(2/3)e^(0.5x) when 0≤x≤2
1 when x>/= 2
Then this represents a mixed distribution both discrete and continuous, so if we were to calculate P(X=2), the answer reads 2/3 e^1 which probably means the 2nd interval is being used with 1F(2). I was wondering why this is the case? Shouldn't the P(X=2) be F(2)F(0)?It's not a mixed distribution, it's continuous.
x is defined on the interval [0,2], anything less than 0 will have F(x)=0, and anything greater than 2 will have F(x)=1.
P(X=2) = 0
You're looking at a cdf, not a pdf.Last edited by ghostwalker; 22032013 at 17:54. 
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 22032013 17:39
(Original post by ghostwalker)
It's not a mixed distribution, it's continuous.
x is defined on the interval [0,2], anything less than 0 will have F(x)=0, and anything greater than 2 will have F(x)=1.
P(X=2) = 0
You're looking at a cdf, not a pdf.
^ where I replaced W with X 
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 22032013 17:52
(Original post by zomgleh)
It says on the lecture slides that it's a mixed distribution though?
^ where I replaced W with X
I'll have another look.Last edited by ghostwalker; 22032013 at 17:54. 
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 22032013 18:01
(Original post by zomgleh)
...
I.e. as d goes to zero. 
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 22032013 18:54
(Original post by ghostwalker)
My apologies. I assumed just glancing at the cdf, that it was continuous at x=0 and x=2.
I'll have another look..
Also, not sure what you mean when you use these limits:
I.e. as d goes to zero
could you please show me how the 2/3 e^1 is obtained? 
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 22032013 19:46
(Original post by zomgleh)
No problem. I'm actually a bit unclear about how to recognize whether it is continuous or discrete just looking at the CDF is the only way to know this through actually sketching it? since then we can identify step functions from smooth ones?
Also, not sure what you mean when you use these limits:
I.e. as d goes to zero
could you please show me how the 2/3 e^1 is obtained?
So, we have 1  (1(2/3)e^(1)) = 2/3 e^1
PS: This is the first time I've seen a question on here about a mixed distribution.Last edited by ghostwalker; 22032013 at 19:47. 
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 22032013 20:55
(Original post by ghostwalker)
For the F(2d), I'm looking at F(x) as x goes to 2, and since the mid portion of the definition is continuous, this is actually 1(2/3)e^(0.5x) evaluated at x=2
This is F(2) which is 1, minus the limit thing I was refering to, i.e. 1(2/3)e^(0.5x) when x=2,
So, we have 1  (1(2/3)e^(1)) = 2/3 e^1
PS: This is the first time I've seen a question on here about a mixed distribution. 
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 22032013 21:03
(Original post by zomgleh)
Thanks a lot, I think I get it...except why is F(2)=1? The CDF says: 1 when x>/= 2.
So, F(2)=F(3)=F(pi)=F(1000000) = 1
Not sure if that's captured what you're getting at. ?
But doesn't F(2) by definition mean P(X<=2) ? 
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 22032013 22:09
(Original post by ghostwalker)
The pdf would be zero for all values greater than 2. (Ignoring esoteric arguments)
So, F(2)=F(3)=F(pi)=F(1000000) = 1
Not sure if that's captured what you're getting at. ?
Yes.
I'm just confused why for a CDF as
1(2/3)e^(0.5x) when 0≤x≤2
1 when x>/= 2
when we want F(2), which = P(X<=2), why don't we choose '1(2/3)e^(0.5x)' instead of just 1? 
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 23032013 03:23
(Original post by zomgleh)
I'm just confused why for a CDF as
1(2/3)e^(0.5x) when 0≤x≤2
1 when x>/= 2
when we want F(2), which = P(X<=2), why don't we choose '1(2/3)e^(0.5x)' instead of just 1?
Where did you get the definitions of the cdf from, in particular the intervals, as they're not in the scan/photo you posted? 
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 23032013 22:30
(Original post by ghostwalker)
Good question.
Where did you get the definitions of the cdf from, in particular the intervals, as they're not in the scan/photo you posted?
It's easy to calculate the P(X=2) from the graph, because it's just the 'jump' of the graph from the smooth part of the curve to the discrete part (step). But I'd like to know how to calculate it without drawing/using the graph. 
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 24032013 02:30
(Original post by zomgleh)
From the same problem, I just typed it out instead (using X instead of W for the random variable), attaching an image though.
'1(2/3)e^(0.5x)' is only valid on the halfopen interval [0,2) which does not include 2. That's why you don't use it to work out F(2).
It's easy to calculate the P(X=2) from the graph, because it's just the 'jump' of the graph from the smooth part of the curve to the discrete part (step). But I'd like to know how to calculate it without drawing/using the graph.Last edited by ghostwalker; 24032013 at 02:32. 
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 24032013 17:15
(Original post by ghostwalker)
For the F(2d), I'm looking at F(x) as x goes to 2, and since the mid portion of the definition is continuous, this is actually 1(2/3)e^(0.5x) evaluated at x=2
This is F(2) which is 1, minus the limit thing I was refering to, i.e. 1(2/3)e^(0.5x) when x=2,
So, we have 1  (1(2/3)e^(1)) = 2/3 e^1
PS: This is the first time I've seen a question on here about a mixed distribution. 
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 24032013 17:29
(Original post by zomgleh)
But how is F(2) = 1 the limit thing? As in, don't we just do that when we're getting the P(X> or >/= x) ?
That was in response to "could you please show me how the 2/3 e^1 is obtained"
I'm not saying "F(2) = 1 the limit thing"
I was saying F(2) is 1. And then minus the limit thing as part of working out P(X=2).
Putting it slightly differently.
P(X=2)= P(X<=2)P(X<2)
=F(2)  limit thing
=1  limit thing. 
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 25032013 20:48
(Original post by ghostwalker)
I think you've misinterpreted what I was saying.
That was in response to "could you please show me how the 2/3 e^1 is obtained"
I'm not saying "F(2) = 1 the limit thing"
I was saying F(2) is 1. And then minus the limit thing as part of working out P(X=2).
Putting it slightly differently.
P(X=2)= P(X<=2)P(X<2)
=F(2)  limit thing
=1  limit thing.
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