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## Statistics and probability

### Course: Statistics and probability > Unit 9

Lesson 7: Geometric random variables- Geometric random variables introduction
- Binomial vs. geometric random variables
- Probability for a geometric random variable
- Geometric probability
- Cumulative geometric probability (greater than a value)
- Cumulative geometric probability (less than a value)
- TI-84 geometpdf and geometcdf functions
- Cumulative geometric probability
- Proof of expected value of geometric random variable

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# Proof of expected value of geometric random variable

Proof of expected value of geometric random variable.

## Want to join the conversation?

- Why are there multipliers next to the expected values? EX: P(x=2) * 2(16 votes)
- You may want to check out the video as a warm up: https://www.khanacademy.org/math/ap-statistics/random-variables-ap/modal/v/expected-value-of-a-discrete-random-variable

Each term in E(X) is`P(X = k) * k`

, where`P(X = k)`

is the weight and`k`

is the number of trials for that term.(10 votes)

- 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.(16 votes)

- 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).(7 votes)
- 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.(4 votes)

- How do you interpret the expected value in this context?(3 votes)
- 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.(5 votes)

- (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)(4 votes)
- 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.(2 votes)

- 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.

You may round your answer to the nearest hundredth.

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)(0 votes)- 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(3 votes)

- one question.

I want to know why it should multiply coefficient in front of each one.

E.G: ""2""**P**(1-P)+ ""3""**P**(1-P)^2

That is P+2*.....+3*....+4*......

Thanks for the help!(1 vote)

## Video transcript

- [Instructor] So right
here we have a classic geometric random variable. We're defining it as the
number of independent trials we need to get a success where
the probability of success for each trial is lowercase p and we have seen this before
when we introduced ourselves to geometric random variables. Now, the goal of this
video is to think about well what is the expected value of a geometric random variable like this and I'll tell you the answer, in future videos we
will apply this formula, but in this video we're actually going to prove it to ourselves mathematically. But the expected value of
a geometric random variable is gonna be one over the
probability of success on any given trial. So now let's prove it to ourselves. So the expected value
of any random variable is just going to be the
probability weighted outcomes that you could have. So you could say it is the probability... The probability that our
random variable is equal to one times one plus the probability that our random variable
is equal to two times two plus and you get the general idea. It goes on and on and on and a geometric random
variable it can only take on values one, two, three,
four, so forth and so on. It will not take on the value zero because you cannot have a success if you have not had a trial yet. But what is this going to be equal to? Well, this is going to be equal to, what's the probability
that we have a success on our first trial? And actually let me
just write it over here. So this is going to be P. What is this going to be? What is the probability
that we don't have a success on our first trial, but we
have one on our second trial? Well, this is going to be one minus p, that's the first trial where
we don't have a success, times a success on the second trial and actually let me do
a few more terms here. So let me erase this a little
bit, do a few more terms. So this is going to be the
probability that X equals two. Sorry, the probability that
X equals three times three and we're gonna keep
going on and on and on. Well, what's this going to be? Well, the probability that X equals three is we're gonna have to get
two unsuccessful trials and so the probability of
two unsuccessful trials is one minus P squared and then one successful
trial just like that. So you get the general idea. If I wanted to rewrite this
and I'm just gonna rewrite it to make it a little bit simpler. So the expected, at least for
the purposes of this proof, so the expected value of X is equal to, I'll write this as 1p
plus 2p times one minus p plus 3p times one minus p squared and we're gonna keep
going on and on and on forever like that. So how do we figure out this sum? And now I'm going to do a little bit of mathematical trickery or
gymnastics, but it's all valid and if any of ya'll have seen the proof of taking an infinite geometric series, then we're gonna do a
very similar technique. What I'm gonna do here
is I'm gonna think about well what is one minus p
times this expected value? So let's do that. So if I say one minus p times
the expected value of X, what is that going to be equal to? Well, I would multiply
every one of these terms times one minus p. So one p times one minus p would be 1p times one minus p. You would get that right over there. What about 2p times one minus p? What would that be equal to? Well, that would be 2p times one minus p and now we're gonna multiply
it by one minus p again. So you're gonna get one minus p squared and so I think you see where this is going and we're just gonna
keep adding and adding and adding from there. Now we're gonna do something
really fun and interesting, at least from a
mathematical point of view. If this is equal to that, if the left-hand side is
equal to the right-hand side, let's just subtract this
value from both sides. So on the left-hand side I would have the expected value of X, that's that, minus this, minus one minus p times the expected value of X. So I'm just subtracting
this from that side, but let me subtract this from that side. Well, I could subtract
this expression from that, but this is equivalent, so I'm just gonna subtract this from that and so what do I get? Well, let's see. I'm gonna have one minus p and then if I subtract
1p times one minus p from 2p times one minus p, well I'm just going to
be left with plus 1p times one minus p and then if I subtract this from that, I'm gonna be left with 1p
times one minus p squared and we're just gonna keep
going on and on and on and so let me simplify this a little bit. If I distribute this negative, this could be plus and then
this would be p minus one and then if we distribute
this expected value of X, we get on the left-hand side the, and let me scroll up a little bit. I don't want to scrunch it too much. So let's see, we have
the expected value of X and then plus p times
the expected value of X. P times the expected value of X minus the expected value of X, these cancel out, is going to be equal to p
plus p times one minus p plus p times one minus p squared and it's gonna keep
going on and on and on. Well, on the left-hand side all I have is a p times expected value of X. If I want to solve for
the expected value of X, I just divide both sides by p. So I get and this is kind of neat through this mathematical gymnastics, I now have, I'm just dividing
everything by p, both sides, on the left-hand side I just
have the expected value of X. If I divide all of these terms by p, this first term becomes one, the second term becomes one minus p, this third term, if I divide by p, becomes plus one minus p
squared, so forth and so on. Now what's cool about this, this is a classic geometric series with a common ratio of one minus p and if that term is
completely unfamiliar to you, I encourage you and this
is why it's actually called a geometric, one of the reasons, arguments for why it's called
a geometric random variable, but I encourage you to review
what a geometric series is on Khan Academy if this
looks completely unfamiliar, but in other places we proved using actually a very similar
technique that we did up here that this sum is going to be equal to one over one minus our common ratio and our common ratio is one minus p. So what is this going to be equal to? And we are really in the
home stretch right over here. This is going to be equal to
one over one minus one plus p. One minus one plus p. Which is indeed equal to one over p. So there you have it, we
have proven to ourselves that the expected value of
a geometric random variable using some, I think, cool mathematics is indeed equal to one over p.