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# Proof of expected value of geometric random variable

Proof of expected value of geometric random variable.

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• Why are there multipliers next to the expected values? EX: P(x=2) * 2
• My teacher tought us that the expected value of a geometric random variable is q/p (where q = 1 - p). I found both formulas on the internet. Which one is the correct one?
(1 vote)
• It depends on how you've set up the geometric random variable. Here, Sal is setting X to be the number of trials you need before you get a successful outcome.
Your teacher, on the other hand, set X to be the number of failures before the first successful outcome.
• When Sal said that E(x) = 1/1-(1-p), I understand how you can get the denominator using the finite geometric series proof he showed on a previous video, but how do you get the one on the numerator? I get 1-(1-p)^n / 1-(1-p).
• Not 1 actually.
The sum of an infinite geometric series is a/1-r ; where a is the first term and r is the common ratio.
Here, "a" happened to be 1 co-incidentally.
• How do you interpret the expected value in this context?
• In this context, X is the number of trials it takes until you get your first success. Therefore, E(X) is the average number of trials it takes to get your first success. With this in mind, it makes intuitively sense that E(X) = 1/p.
• (P) is the average success rate (proportion) of any trial, and a geometric random variable (X) is the number of trials until we reach the first success, so the expected value of (X) should be the number of (P)'s that get us to 1. How many (p)'s are there in 1 ? It's ( 1/p ), which is the expected number of trials until we reach the first success or E(X)
• is the number of (p)'s the same as the value of p?
(1 vote)
• In the video he is referring to, I think he proved that the sum of a geometric series is a(1-r^n)/(1-r), not 1/(1-r). I have watched all his videos until this point. Did I forget something? When did he prove that the sum of a geometric series is 1/1-r?
(1 vote)
• This requires you to know about limit (Or maybe think about it in your mind). 0 < r < 1 for probability.

Let r = 0.5.
Now if n = 1, what is 0.5^n? It's 0.5.
Now if n = 10, what is 0.5^n? It's 9.765625 * 10^-4.
Now if n = 100000, what is 0.5^n? It's a really small number (I can't get the result on my calculator lol). So by this you know when n tends to infinity, r^n tends to 0. Thus a(1 - r^n) / (1 - r) tends to a / (1 - r).
And the first term is 1, so a = 1.

This means you have a probability of 1 / (1 - r) to success 1 time, if each trial has a success probability of r.
• When you subtract (1-p)E(x) from E(x), are you allowed to leave the first term--1p--alone, instead just subtracting the terms of (1-p)E(x) from the corresponding next term in E(x), and still come up with the right answer? I know it's an infinite sequence, so you can't add the unaltered ∞p*(1-p)^∞. Does that mean it doesn't matter, as both sequences go on forever, or does it still change the result?
(1 vote)
• My question is how do we know that we have to take a certain approach while proving the E(X)? In the binomial proof Sal took a different path than the one he took in this video but there also I thought why this path and not something else?
(1 vote)
• For this question:
Lilyana runs a cake decorating business, for which 10% of her orders come over the telephone. Let C be the number of cake orders Lilyana receives in a month until she first gets an order over the telephone. Assume the method of placing each cake order is independent.
Find the probability that it takes fewer than 5 orders for Lilyana to get her first telephone order of the month.

I understand why this is the formula for the answer:
1 - 0.9^4

But can someone explain why this doesn't work?
(4!/(1!3!)*.1^1*.9^3)
• 4!∕(1!3!) × 0.1¹ × 0.9³
only accounts for the possibility of having exactly 1 telephone order among the first four orders, but really we could have 1, 2, 3, or 4 phone orders, and accounting for all those possibilities we get

4!∕(1!3!) × 0.1¹ × 0.9³ +
+ 4!∕(2!2!) × 0.1² × 0.9² +
+ 4!∕(3!1!) × 0.1³ × 0.9¹ +
+ 4!∕(4!0!) × 0.1⁴ × 0.9⁰