WILL MEDAL AND FAN A LOT! Geometric Distribution Help Probability!
@abb0t @Australopithecus @Asevilla5 @arilove1d @Algorithmic @friendzone @e.mccormick @mathstudent55 @mathgeek27 HELP ME PLEASE
Hello!
Instead of tagging incessantly, try posting the question please.
Ok fine xD
Let X be the number of days between the time you purchase a stock and the stock price rises. X is discrete variable with the probability distribution \[f(x)=(\frac{ 1 }{ 2 })^x, (=0 otherwise) x=1, 2, .....\] Find the probability that the stock price decreases and then increases.
Please hurry... Tnx! :)
From the given distribution, you should know that at any time, there is a \(p=\dfrac{1}{2}\) probability of the stock price changing. Now the probability you want to find (that the price decreases then increases) is essentially the same as determining the probability to the first success after one failure, if we consider an increase in price a success (in this case it doesn't matter whether we pick an increase or decrease as a success because the complement event has the same probability). This probability is given by the geometric expression, \[(1-p)^{2-1}p=\left(1-\frac{1}{2}\right)^1\frac{1}{2}=\frac{1}{4}\]
This assuming there are only two exchanges being considered (a decrease in price followed by an increase). It could be that the question wants a general expression for an increase in price following \(k\) decreases, in which case you would simply replace \(2\) in the exponent with \(k\).
The book says If the stock price decreases then increases, then x =2. Now p{x=2} = (1/2)^2=1/4
What does it mean x=2?
\(x=2\) must refer to the number of price changes.
Oh. What about the chance of no success? Where is that in the equation thing?
In this case, it's the same. \(p\) is the chance of success, and \(1-p\) is the chance of failure. In this case, the success prob is \(\dfrac{1}{2}\), so \(p=1-p\).
Woah. Ok. Thanks
yw
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