Converting Prior Knowledge into a Bayesian Prior Suppose I flip a coin 10 times and get 7 heads - based on a Binomial Distribution, I can create a Likelihood Function and estimate the probability of getting a head is 0.7.
Now, suppose someone tells me that they have very strong evidence to believe that the probability of getting a head is 0.9. Suppose I decide to use a Beta Distribution(alpha, beta) as a prior.
My Question: I know that the Expected Value of the Beta Distribution is 'alpha/(alpha + beta)'. If I want to use the Prior Information that was provided to me - does this mean that I have choose values of 'alpha and beta' such that 'alpha/(alpha+beta)' = 0.9? For example - based on the n = 10 samples I observed, in this case, would alpha = 9 and beta = 1?
 A: If I understand correctly, your problem is not defined correctly:

*

*The likelihood is a Binomial distribution (you flip the coin)

*The Prior is a Beta distribution (representing the distribution of
the parameter p of the Binomial)

*Beta is a conjugate prior for Binomial, it means the posterior is still a Beta distribution.

Here you can find the following details:

Therefore, in order to identify the "believe", that is a Beta distributions, you need 2 parameters:

*

*alpha & beta (your problem is solved)

OR

*

*mean & variance, in order to solve this equation:

$$\mathbf E[X]=\frac\alpha{\alpha+\beta}$$
$$\mathbf V[X]= \frac{\alpha\beta}{(\alpha+\beta) ^2(\alpha+\beta+1) }$$
as reported here: Calculating the parameters of a Beta distribution using the mean and variance.
Therefore: Variance is missing.
A: 
does this mean that I have choose values of 'alpha and beta' such that 'alpha/(alpha+beta)' = 0.9?

That's correct, but the actual values you choose determine how strong your prior is.
In this particular setup, there's an intuitive explanation of the Beta prior: it is equivalent to having previously observed $\alpha$ heads and $\beta$ tails from the same coin.
You can thus see that $(\alpha, \beta)=(9,1)$ is much weaker than $(\alpha, \beta)=(900000,100000)$.
A: The typical approach in this case would be to use the conjugate prior (the beta distribution) for the unknown probability parameter and parameterise this using a stipulated prior mean $0 \leqslant \mu_0 \leqslant 1$ and prior effective-sample-size $n_0 \geqslant 0$.  You can then use the reparameterisation of the beta distribution with parameters:
$$\begin{align}
\alpha &= \mu_0 (n_0+2), \\[6pt]
\beta &= (1-\mu_0) (n_0+2). \\[6pt]
\end{align}$$
From the information you have given you would choose $\mu_0 = 0.9$ and you would then have to choose an appropriate value for the prior effective sample-size.  This is a measure of the strength of your prior belief --- the higher the effective-sample-size the stronger your prior belief (and so the lower the prior variance).  You describe the prior belief as "very strong" so you could choose a value like $n_0 = 100$ to represent this, which would mean that you consider your prior belief to be as strong as if it were formed from one-hundred observed data points.
