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Suppose I have a random variable:

\begin{equation} X \sim \begin{cases} 1 \text{ with probability } p \\ y \sim \mathit{unif}(1,k) \text{ with probability } 1 - p \end{cases} \end{equation}

Where $\mathit{unif}$ is the discrete uniform distribution. I am trying to compute the expected value $E[X].$ I think it should just be simply:

\begin{equation} 1 \cdot p + E[y] \cdot (1-p) = 1 \cdot p + (1 + k)/2 \cdot (1-p) \end{equation}

But I am not sure if I am overlooking something.

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    $\begingroup$ The given answers are both correct for calculating the answer in your concrete problem. I would add, however, that in a general case, you would apply conditional probabilities. For discrete distributions, and naming the random variable that $X$ depends on $Z$, we get: $$\mathbb{P}(X = k)=\sum_{z\in\text{supp}(Z)} \mathbb{P}(X = k | Z = z) \mathbb{P}(Z=z)$$ (law of total probability). This you can then evaluate further (in this concrete example, of course, the sum is only over two values), and in particular evaluate the expectation via $\sum_k k \mathbb{P}(X=k)$. $\endgroup$ Commented Nov 17, 2022 at 15:25
  • $\begingroup$ Hi. I'm used to notation $y \sim \mathit{unif}(\text{some set})$ for the uniform distribution on a given set, but I've never encountered notation $y \sim \mathit{unif}(1,k)$. I don't know what the $1$ and $k$ parameters mean in your case. Could you please explain? From the expected value calculation I can guess that it's maybe the uniform distribution on closed segment $\left[1,k\right]$, or on open segment $\left]1,k\right[$, or on the two-element set $\{1, k\}$, or something else... $\endgroup$
    – Stef
    Commented Nov 18, 2022 at 9:08

2 Answers 2

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Your random variable takes the value $1$ with probability $p+\frac{1-p}{k}$, and takes each value $j\in\{2, \dots, k\}$ with probability $\frac{1-p}{k}$. So the expectation is simply $$ \begin{align*} EX = & 1\times \big(p+\frac{1-p}{k}\big) + \sum_{j=2}^k j\times\frac{1-p}{k} \\ = & p+\frac{1-p}{k}\times \sum_{j=1}^k j \\ = & p+\frac{1-p}{k}\times\frac{k(k+1)}{2} \\ = & p+\frac{(1-p)(k+1)}{2}. \end{align*}$$

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Your answer is okay and is justified below.

You can write:$$X=B+(1-B)Y$$where $B\sim\text{Bernoulli}(p)$ and $Y\sim\text{Unif}(k)$ are independent random variables.

Then:$$\mathbb EX=\mathbb EB+\mathbb E[(1-B)Y]=\mathbb EB+\mathbb E(1-B)\mathbb EY=p+(1-p)\left(\frac12+\frac12k\right)$$

This way of working is recommendable in situations where the rv is defined by means of cases.

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