# Clarification on variance and expected value problem

Suppose that X is a random variable where:

$P(X = 1)$ = 1/2

$P(X = 2) =$1/4

$P(X = 4)$ = 1/4

Suppose Y is another random variable that takes values from the set $$Y = {{1, 2, 4}}$$

but the probabilities that it takes each value are unknown and some of them could be zero.

What is the largest value that $E(Y)$ and $var(Y)$ can take?

I know that the largest $E(Y) = 4$

I also know that the answer for $var(Y)$ is $2.25$

I just don't know the necessary steps to get to the answer.

Hints

How can you maximize $\mathbb{E}[Y]$? Note that it is given by

$$\mathbb{E}[Y] = \mathbb{P}(Y = 1) \cdot 1 + \mathbb{P}(Y = 2) \cdot 2 + \mathbb{P}(Y = 4) \cdot 4.$$

The variance is given by

$$\operatorname{Var}(Y) = \mathbb{E}[Y^2] - \mathbb{E}[Y]^2.$$

So, what you want to do, is maximize $\mathbb{E}[Y^2] - \mathbb{E}[Y]^2$ which means that you want to try to make $\mathbb{E}[Y^2]$ large and at the same time make $\mathbb{E}[Y]$ small. Remember that

$$\mathbb{E}[Y^2] = \mathbb{P}(Y = 1) \cdot 1^2 + \mathbb{P}(Y = 2) \cdot 2^2 + \mathbb{P}(Y = 4) \cdot 4^2.$$

The expectation is maximized when the probability that the random variable $Y$ attains its largest value is maximized. Since there are no constraints, this is achieved when $Y=4$ with probability $1$ and $Y=1,2$ with probability $0$. Formally, in this case $$E[Y]=1\cdot P(Y=1)+2\cdot P(Y=2)+4\cdot P(Y=4)=0+4\cdot1=4$$

The variance, as a means of dispersion, is maximized when the possible values of $Y$ are distributed as far away from the mean as possible, or in other words when the values of $Y$ are as less concentrated as possible. So, put as much weight on $1$ and $4$ at the same time, which can be done by choosing $$P(Y=1)=P(Y=4)=1/2$$ and $P(Y=2)=0$. In this case $$E[Y]=\frac12\cdot1+\frac12\cdot4=2.5$$ with $E[Y]^2=2.5^2=6.25$ and $$E[Y^2]=\frac12\cdot1^2+\frac12\cdot4^2=8.5$$ So $$Var(Y)=E[Y^2]-E[Y]^2=8.5-6.25=2.25$$