1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

Hints

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

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

The variance is given by

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

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

\begin{equation} \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. \end{equation}

$\endgroup$
1
$\begingroup$

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 $$Var(Y)=E[Y^2]-E[Y]^2=8.5-6.25=2.25$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.