Expected value of binomial distribution

• hotvette
In summary, the homework statement states that a random variable Y has a binomial distribution with n trials and success probability X, where n is a given constant and X is a uniform(0,1) random variable. E[Y] is the expected value of Y, which is np.

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Homework Statement

A random variable Y has a binomial distribution with n trials and success probability X, where n is a given constant and X is a uniform(0,1) random variable. What is E[Y]?

E[Y] = np

The Attempt at a Solution

The key is determining the probability of success, which is stated as X, thus the answer should be nX. But X is a uniform(0,1) random value, which is what I find confusing. My first thought was that X is constant = 1 (because uniform distribution) and therefore E[Y] = nX = n but I don't think that's right (it's like a double headed coin, n flips = n successes). I guess I don't understand how the fact that X is a uniform(0,1) RV enters into the problem. Any hints?

I think this one is simple. If ## X ## is random and uniform between ##0 ## and ##1 ##, that would make ## EX=\frac{1}{2} ##. From there, ## EY ## follows immediately. Perhaps a more detailed calculation is necessary, but I'm thinking it isn't. ## \\ ## Edit: It might pay to consider ## EY ## for ## n=1 ##: ## EY_{n=1}= \int\limits_{0}^{1} p(x) x \, dx ##. And note: ## p(x)=1 ## for ## 0<x<1 ##, (with the uniform distribution), so that ## \int p(x) \, dx=1 ##.

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hotvette said:
it's like a double headed coin, n flips = n successes
Only if X = 1. But X can also be 0, leading to n flips = n misses ...

 Oh boy, Charle is faster, giving it all away, and thhus robs you from the exercise...

Thanks for the replies! I'm not sure I follow, though. I completely understand that E[X ]= 1/2 by using the definition of expected value of a continuous RV. But what I don't understand is why E[Y] = nE[X] (and therefore n/2). Seems to me E[Y] = np = nX from the problem statement. There is something I'm missing conceptually I think.

hotvette said:
Thanks for the replies! I'm not sure I follow, though. I completely understand that E[X ]= 1/2 by using the definition of expected value of a continuous RV. But what I don't understand is why E[Y] = nE[X] (and therefore n/2). Seems to me E[Y] = np = nX from the problem statement. There is something I'm missing conceptually I think.
That is only the expectancy once you know what ## p ## is going to be given to you, basically by spinning a wheel before the experiment, and where the dial lands tells how much the coin will be biased in the ## n ## trials. ## \\ ## In any case, for the complete expectancy of ## Y ##, they are asking, what can you expect to get on the average?, given we are first going to spin the wheel, and then bias the coin accordingly. ## \\ ## I computed ## EY_{n=1} ## =for a single trial. If ## Y=X_1+X_2+...+X_n ##, then ## EY=n \, EX ##.

OK, I think I understand. Thanks!