# Binomial distribution with different success probabilities

#### Buckeye

##### Member
I have n independent trials with different success probabilities θi for i=1 to i=n
I want to show that...

μ=nθ with θ= (1/n) ∑θi for i=1 to i=n

I guess this makes sense intuitively for θ to be the average of all θi. Can someone help show this? The proof for a constant θ is straight forward. I'll leave the variance for another day.

Edit: To be clear, this is not a homework exercise. I found it in my book while reviewing the material. I attached the problem below.

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#### Dason

I'm not clear what your question is exactly.

#### Buckeye

##### Member
I apologize. I'm new to this website. The attached problem is courtesy of John E. Freund's Mathematical Statistics with Applications eighth edition. It's an undergraduate text. I found the problem interesting because from what I know of the binomial distribution, the success probability is held constant.

#### Dason

Define $$X_i$$ to be the outcome of the ith trial. Then $$X = \sum_{i=1}^n X_i$$.
Use that to rewrite $$\mu_X = E[X]$$ by inserting the definition of X into the expected value. See if you can break that up. Keep in mind that $$E[X_i] = \theta_i$$