Proof of expected value of geometric random variable | AP Statistics | Khan Academy

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. 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? 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. The expected value of a geometric random variable is going to be 1 over the probability of success on any given trial.

So now let's prove it to ourselves. 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 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.

A geometric random variable can only take on values 1, 2, 3, 4, so forth and so on. It will not take on the value 0 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 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. 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 2. Sorry, the probability that x equals 3 times 3, and we're going to keep going on and on and on. Well, what's this going to be? Well, the probability that x equals 3 is we're going to have to get two unsuccessful trials, and so the probability of two unsuccessful trials is (1 minus p) squared, and then one successful trial just like that.

So you get the general idea. If I wanted to rewrite this, I'm just going to rewrite it to make it a little bit simpler. So the expected, at least for the purposes of this proof, the expected value of x is equal to... I’ll write this as 1p plus 2p*(1 minus p) plus 3p(1 minus p squared), and we're going to keep going on and on and on forever like that.

So how do we figure out this sum? Now I'm going to do a little bit of mathematical trickery or gymnastics, but it's all valid, and if any of you all have seen the proof of taking an infinite geometric series, then we're going to do a very similar technique.

What I'm going to do here is I'm going to think about, well, what is (1 minus p) times this expected value? So let's do that. If I say (1 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 (1 minus p). So 1p times (1 minus p) would be 1p*(1 minus p). You would get that right over there. What about 2p times (1 minus p)? What would that be equal to? Well, that would be 2p*(1 minus p), and now we're going to multiply it by (1 minus p) again, so you're going to get (1 minus p) squared.

And so I think you see where this is going, and we're just going to keep adding and adding and adding from there. Now, we're going to 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 (1 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 going to subtract this from that.

And so what do I get? Well, let's see. I'm going to have (1 minus p), and then if I subtract 1p(1 minus p) from 2p(1 minus p), well, I'm just going to be left with plus 1p(1 minus p). And then if I subtract this from that, I'm going to be left with 1p(1 minus p squared), and we're just going to keep going on and on and on.

So let me simplify this a little bit. If I distribute this negative, this could be plus, and then this would be p minus 1. And then if we distribute this expected value of x, we get on the left-hand side... 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. It is going to be equal to p plus p*(1 minus p) plus p*(1 minus p squared), and it's going to keep going on and on and on.

Well, on the left hand side, all I have is a p times the 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 1. The second term becomes (1 minus p). This third term, if I divide by p, becomes plus (1 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 (1 minus p), and if that term is completely unfamiliar to you, I encourage you... and this is why it's actually called a geometric random variable, 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.

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 1 over (1 minus our common ratio), and our common ratio is (1 minus p). So what is this going to be equal to? We are really in the home stretch right over here. This is going to be equal to 1 over (1 minus 1 plus p), which is indeed equal to 1 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 1 over p.