I've seen a few of these questions of the form $max$(X,constant) but they get more math heavy than I know mine should get. Could someone help me understand how to solve this?
What we are given is that the Distribution of our random variable X is {$1,2,3,4,5$} with probabilites {$1,\frac{39}{51},\frac{26}{50},\frac{13}{49},1$} respectively. Given this, I am told to find $E(X,3)$. How would I do this? I know there shouldn't be any integrals involved.
Let us define $Z:=\max\left\{X,3\right\}.$ Now, first let us consider the values of $Z$ conditioned on the values of $X$. That is, it should be clear that $$[Z|X=x] = \left\{\begin{matrix} 3, & x = 1,\\ 3, & x =2,\\ 3, & x = 3,\\ 4, & x = 4,\\ 5, & x = 5. \end{matrix}\right.$$
Now, because $[Z|X=x]$ is always constant, it's expectation is simply its constant value, i.e., $$[Z|X=2] = 3 \Rightarrow \mathbb{E}(Z|X=2) = 3.$$ Therefore, by conditioning we can simply compute the expectation in the normal way, i.e., $$\mathbb{E}(Z) = \sum_{x=1}^{5}\mathbb{E}(Z|X = x)\mathbb{P}(X=x) \\ = 3\left(\mathbb{P}(X=1) + \mathbb{P}(X=2) + \mathbb{P}(X=3)\right) + 4\mathbb{P}(X=4) + 5\mathbb{P}(X=5).$$
Where you can insert the PMF appropriately (it seems like in the question there is a problem with the PMF you gave as it doesn't sum to 1).
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A simpler method would be to use the following fact: for a discrete random variable $X$ and function $f$
In our case, $f(X)=\max \{X,3\}$, thus the solution is $$\mathbb{E}(\max \{X,3\}) =\sum_{x=2}^5\max\{x,3\}\mathbb{P}(X=x).$$ The technique in the original answer is more general though and will aid in getting the expectation of more complicated random variables, e.g., $\max\{X,Y\}$ where $X$ and $Y$ are both random variables.