Stats Question

[128nigel]

The number of days a particular stock increases in a given five-day period is uniformly distributed between zero and five inclusive. In a given five-day trading week, what is the probability that the stock will increase exactly three days?
A) 0.333.
B) 0.200.
C) 0.167.
D) 0.600.

 

[cpk123]

3/5 = 0.6 Choice D
Uniform distribution –> .2 probability any day.

 

[128nigel]

Thanks!! Thats what i figured

 

[malnoll]

0-5 inclusive makes it 6 possibilities

 

[cpk123]

but the probability would be
(3 - 0) / (5-0)

 

[malnoll]

The one topic I did bad on June test was Quant….but I figured you could have 0days, 1day…..5days that the stock could increase. Anyone want to explain that to me, this is exactly why I did bad on quant

 

[128nigel]

We were all wrong according to Qbank…
Answer:
If the possible outcomes are X0,1,2,3,4,5), then the probability of each of the six outcomes is 1/6=0.167.
… shouldnt the “3 days” part of the question come into play?

 

[128nigel]

nvm. dumb question
X= 3
P(3) = 1/6

 

[dinesh.sundrani]

6C3 * (0.166667)^3 * (0.8333333)^3
20* 0.0046296324 * 0.5787036 = 0.05358 = What have I done wrong guys??

 

[LongOnCFA]

dinesh,
the question says the # of days up are uniformly distrbuted, not the probability that it will go up on any given day.

 

[dinesh.sundrani]

still didn’t understand… could you elaborate more?

 

[cpk123]

since it is # of days is uniformly distributed –>
P(3) = 1/6 = (also equal to P(1), P(2) or so on).
so you need 1/6 to be the answer to the question asked…
English, TOEFL, $$%*(#%$(#%( etc. etc. etc.

 

[malnoll]

isn’t the assumption that on any day, it is 50/50 for a stock to go up? all you would have to to do is find the probability that it goes up exactly 3 days…..or is that wrong

 

[ancientmtk]

Who ever wrote this needs to take ESL again

 

[JoeyDVivre]

Actually, they may need to take stats class. So for a cookie prove that if “number of days a particular stock increases in a given five-day period is uniformly distributed between zero and five inclusive” then the outcome of day x and day y are not independent and thus EMH is disproved.

 

[malnoll]

Full for a cookie, but the answer/non-sarcastic explanation would be good Mssr. JoeyD

 

[JoeyDVivre]

Well cpk would be right on the question as written. If it’s uniform that means if X = # days stock is up then P(X=3) = P(X=4) = P(X=1) = etc = 1/6
But if the days are independent and the probability of up is fixed at p, the number of days the stock is up is binomial not uniform.
To do a proof that it can’t be uniform, just look at 2 days and write the probabilities if they are independent and then show that there is no solution. It then generalizes.
A more general and pretty deep result is that if X + Y is uniform then X and Y cannot be independent no matter what the distribution of X or Y and they can be of completely different distributions. Tough to prove though.

 

[malnoll]

thanks