Expected value of binomial distribution

[hotvette]

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]?

Homework Equations

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?

 

[Charles Link]

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 $\int p(x) \, dx=1$.

 

[BvU]

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 ...

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

 

[hotvette]

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.

 

[Charles Link]

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 $EY=n \, EX$.

 

[hotvette]

OK, I think I understand. Thanks!