Random Variables - Expected value and Variance

Question

Let X be the number of heads obtained in 2 independent tosses of a fair coin.

a) find $\mathbb E(1/(1+X))$

I began by making a table to show that probability distribution. X can be 0, 1 or 2 and the corresponding probabilities are 0.25, 0.5 and 0.25.

IF I use the formula $\Bbb E(Y) = \Bbb E(g(X)) = \sum_x g(x)\Bbb P(X=x)$
[i'm presuming this is a discrete random variable]

what can I do next?

Answer 1

IF I use the formula $\Bbb E(Y) = \Bbb E(g(X)) = \sum_x g(x)\Bbb P(X=x)$
[i'm presuming this is a discrete random variable]

You don't need to presume it is a discrete random variable.   You had just shown this to be so when you calculated the probability mass function of $X$.$$\Bbb P(X=x)=\begin{cases}1/4& :& x=1 ~\lor~ x=3\\ 1/2 &:& x=2\\0&:& \text{otherwise}\end{cases}$$

Hence using the formula: $\Bbb E(g(X)) = \tfrac 14 g(1)+\tfrac 12 g(2)+\tfrac 14g(3)$

Next, evaluate $g(\;)$ at $1,2,3$.

Answer 2

Evaluating $g(x)$ atthe possible $x=0,1,2$ values yield $g(x)=1, \frac{1}{2}, \frac{1}{3}$.

Hence, $$\mathsf{E}(\frac{1}{1+x})=1(0.25)+\frac{1}{2}(0.5)+\frac{1}{3}(0.25)$$