Solved – Expected value of coin tossed twice

expected valuerandom variable

Here's the problem:

A (Laplace) coin with sides $0$ and $1$ is tossed twice, each time independently from the other. Let

$f:=$ maximum of both results

$g:=$ sum of both results.

What is the expected value and variance of $f$ and $g$?

I think $E[g]=2*\frac{1}{4} + 1*\frac{1}{2}+0*\frac{1}{2}+0*\frac{1}{4}=1$, but I'm not sure how to go about calculating the rest. Any guidance would be appreciated!

Best Answer

Only two coins, so it's easy to see all combinations, each with 1/4 probability:

toss  f  g
00    0  0   
01    1  1
10    1  1
11    1  2

Expected value of random variable $X$ denoted as $E[X]$ is a sum of results, multiplied by their probabilities. I.e.,

$E[g] = \frac{1}{4}*0+\frac{1}{4}*1+\frac{1}{4}*1+\frac{1}{4}*2$

Now, variance is

$E[X^2] - (E[X])^2$

I believe it is an easy task to go from these definitions to the answer of your question.

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