It’s been a frustrating week to say the least. I think I’ve mentioned before that I live in Toronto, but I may not have mentioned that I live right beside the biggest park in the city, High Park. I love the park but this time of year there is a plant called Golden Rod that blooms which gives me AWFUL allergy symptoms. I spent all of last week itching my eyes and blowing my nose. I tried using an antihistamine which helped initially but then it started to lose its’ effectiveness. Trying to get work done while constantly blowing nose was a huge struggle and I didn’t end up getting through the unit Random Variables.

Right now I’m trying to remind myself that sometimes we have days/weeks where things don’t go according to plan due to circumstances that are out of our control. When this happens, it’s best to reassess your goals and do your best to simply do the best you can under the circumstances. I’m trying to see it that way now but I’m still having a hard time not feeling frustrated. BUT, I just realized now looking back through my notes that I actually wrote out 35 pages of notes in my notebook so, even though I didn’t get through the unit, I at least out forward a solid effort.

Ok, now that I got that little pity party out of the way… The week started off with what’s known as the Bernoulli Distribution. As I write this, I’m realizing now that I still don’t have a great idea of what this principle is and how it works. Here are the notes I took that explains it:

As I said, I still don’t really understand this principle but it seems to me that it’s used when 1) you’re trying to find the mean of two differently weighted outcomes (i.e. p and 1 – p) and 2) to then find the standard deviation of that distribution. It seems it can only be used when trying to find the mean and S.D. of a situation where there are only two outcomes (described as either ‘success’ of ‘failure’) but can go through n number of trials. When there are n trials, to find the mean you simply use the equation n * p and to find the S.D. you would use the equation √ n * p * (1 – p). For example:

• Q. There are x number of employees at a company, 40% of which are men. The CEO chooses 8 at random to take for lunch. What is the mean and S.D. of the men he will choose?
  • μ = n * p
    • = 8 * .4
    • = 3.2
  •  σ = √ n * p * (1 – p)
    • = √ 8 * .4 * (1 – .4)
    • = √ 8 * .4 * .6
    • = √ 1.92
    • = ~1.39

The next thing I learned about was what a geometric variable is and the difference between it and a binomial variable. Here’s a chart that highlights the only difference between the two:

	Binomial Variable	Geometric Variable
‘Success’ or ‘Failure’	✓	✓
Independent Trials	✓	✓
Fixed Number of Trials	✓
Same Probability per Trial	✓	✓

To reiterate the above, for a variable to be either Binomial or Geometric, it must 1) have two outcomes that could be classified as either a ‘success’ or ‘failure’, 2) each trial must be independent, and 3) the ‘success’ and ‘failure’ must have the same probability of occurring on each trial. The difference between the two, however, is that, where a binomial variable has a set number of trials, the number of trials of a geometric variable could theoretically go on indefinitely. One way to think about it would be that with a binomial variable you’d ask yourself, “how many successes in x trials?” whereas with a geometric variable you would ask, “how many trials until success?”.

Going through a number of questions on random variables and their expected values, I eventually came up with a method to determine the E.V. of each possible weighted outcome. Here’s a photo from my notes that shows two examples of the tables I made for different questions (with 2 tables used in each question):

In the bottom example, the question I was asked was something along the lines of:

• Q. Your friend asks you to play one of two betting games. In the first game you flip a fair coin 4 times and if you get 2-or-less heads you win $1 but if you get 3-or-more heads you lose $3. In the second game you flip a fair coin 5 times and if you get all 5 heads you win $100 but if you get 4-or-less heads you lose $4. What is the expected value of both games and which should you choose to play, if any?
• Game 1:
  • P(0, 1, or 2 heads) = [(_4C_0) + (_4C_1) + (_4C_2)] / 2^4
    • = [(1) + (4) + (6)] / 16
    • = 11/16
  • E.V. (0, 1, or 2 heads) = 11/16 * $1 = $0.6875
  • P(3 or 4 heads) = [(_4C_3) + (_4C_4) / 2^4
    • = [(4) + (1)] / 16
    • = 5/16
  • E.V. (3 or 4 heads)  = 5/16 * -$3
    • = -$0.9375
  • Total E.V. = $0.6875 + -$0.9375
    • =-$0.25
• Game 2:
  • P(5 heads) = _5C_5 / 2^5
    • = 1/32
  • E.V. (5 heads) = 1/32 * $100
    • = $3.125
  • P(0, 1, 2, 3, or 4 heads) = [(_5C_0) + (_5C_1) + (_5C_2)  (_5C_3) + (_5C_4)] / 2^5
    • = [(1) + (5) + (10) + (10) + (5)] / 2^5
    •  = 31/32
  • E.V. (0, 1, 2, 3, or 4 heads) = 31/32 * -$4
    • = -$3.875
  • Total E.V. = $3.125 + -$3.875
    • = -$0.75
• Therefore, since the total E.V. of Game 1 is -$0.25 and the total E.V. of Game 2 is -$0.75, Game 1 would be the better option to play, however it would be best not to play either since they both have a negative expected value.

By making the tables featured above, it made it much easier for me to break a question down into individual parts and to keep track of the different values. Being that each question required quite a few steps to get to the right answer, I found keeping everything neat and organized in a table made these types of questions much easier to solve.

To end on a positive note, there were a few things that happened this week that I was happy about. I bought Microsoft Word and Excel for my home computer and, although the $190 it cost to buy the programs wasn’t exactly positive, it’s much nicer being able to use my home computer to write these posts on as opposed to my tiny work-laptop I’ve been using until now. As well, even though I only got a pitiful 3/7 questions correct on the unit test before I gave up, 3/4 questions I got wrong was because of simple miscomprehension mistakes, not because I didn’t understand the math. Lastly, this coming week is the last week of my ‘summer-schedule’ where I work during the day. Beginning Week 55, I’ll be back to my normal schedule where I work in the evenings which I find much better suited for me in general but also particularly for me to work through KA. I also think I should be over these dreadful allergies by then, too. Happy days are on the horizon folks! Just need to get through Week 54.