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Expectation of X watch

1. Hey guys, I have a question. I have been given that expectation of X=0.8 and the probability of r=-1,0,1,2 is given by b,c,b,c and I have to calculate the values of b and c. I know i have to multiply the r by the probability but then i would get -1b, 0c, 1b and 2c but wouldnt that mean the b's will cancel out each other. How would i find that out? Thanks in advance
2. i cant remember strictly how to do this but i think i have it:

E[X] = 0.8

P(R=-1) = b
P(R=0) = c
P(R=1) = b
P(R=2) = c

Now... E[X] = -1b + b + 2c = 0.8
which then, as you said, the b's cancel giving 2c=0.8, hence c=0.4

then, as all probabilities, when totalled, equal 1
this means 2b + 2c =1
as you know the value of c, you can work out the value of b

Hope this helps!
3. (Original post by gobbledygook88)
i cant remember strictly how to do this but i think i have it:

E[X] = 0.8

P(R=-1) = b
P(R=0) = c
P(R=1) = b
P(R=2) = c

Now... E[X] = -1b + b + 2c = 0.8
which then, as you said, the b's cancel giving 2c=0.8, hence c=0.4

then, as all probabilities, when totalled, equal 1
this means 2b + 2c =1
as you know the value of c, you can work out the value of b

Hope this helps!
ok thanks i understand but the one thing i dont understand is why have you put 2b? i have no idea where that came from
4. the b's and c's are the probabilities for the given values of r
[if that makes sense... i cant think of another way to rephrase this...i'm in statistics mode at the moment ]

and as there are 2 of each, I have written 2b + 2c = 1 [the total probability]
5. (Original post by gobbledygook88)
the b's and c's are the probabilities for the given values of r
[if that makes sense... i cant think of another way to rephrase this...i'm in statistics mode at the moment ]

and as there are 2 of each, I have written 2b + 2c = 1 [the total probability]
oh ok i think i understand now i.e. the b and c will always have the same coefficient? or no?
6. ...not quite...

you have written that there are 4 discrete values for r: r=-1,0,1,2
and for those 4 values of r, there are 4 respective probabilities P(R=r) = b,c,b,c [respectively]

now, P(R=r) = 1 [as all probabilities total 1]
this means that b + c + b + c = 1
hence, 2b + 2c = 1
the rest then follows from above

[sorry, i meant to say i am stuck in statistics 2 mode...its nice...but confusing at the same time...its better to learn the whole of S1 first...unlike me i forgot most of it...thats why i'm confused]
7. (Original post by gobbledygook88)
...not quite...

you have written that there are 4 discrete values for r: r=-1,0,1,2
and for those 4 values of r, there are 4 respective probabilities P(R=r) = b,c,b,c [respectively]

now, P(R=r) = 1 [as all probabilities total 1]
this means that b + c + b + c = 1
hence, 2b + 2c = 1
the rest then follows from above

[sorry, i meant to say i am stuck in statistics 2 mode...its nice...but confusing at the same time...its better to learn the whole of S1 first...unlike me i forgot most of it...thats why i'm confused]
oh ok i understand it now thanks and if you havent learnt the whole of S1 youre very clever since you are able to help me and thanks again!
8. thats ok...make sure you learn this stuff...not so much combinatorics...cause the other statistics modules are much more theoretical...and if you like that sort of stuff...you'll love it!
9. (Original post by gobbledygook88)
thats ok...make sure you learn this stuff...not so much combinatorics...cause the other statistics modules are much more theoretical...and if you like that sort of stuff...you'll love it!
lol sry but i have one more question it tells me to calculate the standard deviation of X so i know i have to make another row for r^2 x probability but what if the probability of r = 0 is 0.2 doesnt that mean 0x0.2 will equal 0. I know this question might actually be obvious but it just seems strange to me cause there a probility of of r=0 so how can it be 0 by calcualting the standard deviation.
Sorry if it doesnt make any sense/hard to understand ill gladly rewrite it if it doesnt.
10. hehe
you need to remember that r is a discrete random variable, its just a value that happens to occur in an event
now, there will obviously be a probability for this value (r=0) as it is a valid occurrence, it just turns out that r=0, which then means that 0x0.2 =0
its not wrong, its just how it is

you are on the right track with r^2, then you will find Var[X]

hope that clears some things up
11. (Original post by gobbledygook88)
hehe
you need to remember that r is a discrete random variable, its just a value that happens to occur in an event
now, there will obviously be a probability for this value (r=0) as it is a valid occurrence, it just turns out that r=0, which then means that 0x0.2 =0
its not wrong, its just how it is

you are on the right track with r^2, then you will find Var[X]

hope that clears some things up
ok thanks for the help couldnt have done it without you

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