a) In a horserace, W denotes the winnings. The probability of winning 5USD is 0.02, the one for winning 25USD is 0.002 and the one for winning 100USD is 0.0005 (the probability of not winning any money is 0.9775).
For this part I got $E= \frac 1 5$
b)The random variable X has the pdf f : R→R
$$
f(x) :=
\begin{cases}
\frac{4x(9-x^2)}{81}, & \text{for $0\le x\le3$} \\
0, & \text{otherwise}
\end{cases}
$$
I got$ E=1$ here after integration
c) For the random variable Y it remains that $$ P[Y=y] := \begin{cases} \frac{1}{2}^y, & \text{for $y\in (1, 2, 3, 4, 5)$} \\ c, & \text{for $y=6$} \\ \end{cases} $$ with a value of c so that $\sum_y P[Y=y] = 1$ is reached
I am uncertain on how to start with this one, how do I find c and how do I integrate this?
d) The random var. z has a pdf g : R→R $$ g(z) := \begin{cases} \frac {10}{9z^2}, & \text{for $1\le z\le10$} \\ 0, & \text{otherwise} \end{cases} $$
For this part I also got $E=1$
Did you get the same answers as me? I'm not sure if I'm doing it right so I would appreciate some feedback.
a). is correct, but b). and d). are wrong. You need to compute $\int_{\mathbb{R}} xf(x)dx$ to get the expected value, not $\int_{\mathbb{R}} f(x)dx$ (which is always equal to $1$ by definition since $f$ is a probability density function). Same for d).
As for c). you must have that the sum of probabilities is $1$. Therefore, $$ 1 = \sum_{k=1}^6 \Pr[Y=k] = \sum_{k=1}^5 \frac{1}{2^k}+ c $$ which allows you to find $c$ (check that $c\geq 0$, to make sure you did not nake an obvious mistake). Once you have $c$, you can get the expected value as $$ \mathbb{E}[Y] = \sum_{k=1}^6 k\Pr[Y=k] = \sum_{k=1}^5 \frac{k}{2^k}+ 6c\,. $$