Question

Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive constant. 1. find

c. 2. compute p(x > 2). 3. compute e[x] and var(x). 4. let y = x2+1, find the pmf of y: 5. compute

Answer 1

Looks like the PMF is supposed to be


`\mathbb P(X=x)=\begin{cases}\frac c3&\text{for }x\in\{1,5\}\\\\\frac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}`

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:


`\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\frac c3+\frac c6+\frac c3=\frac{5c}6=1\implies c=\frac65`

Next,


`\mathbb P(X>2)=\mathbb P(X=5)=\frac c3=\frac25`


`\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\frac c3+\frac{2c}6+\frac{5c}3=\frac{7c}3=\frac{14}5`


`\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2`

`\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\frac c3+\frac{4c}6+\frac{25c}3=\frac{28c}3=\frac{56}5`

`\implies\mathbb V(X)=\frac{56}5-\left(\frac{14}5\right)^2=(84)/(25)`

If
`Y=X^2+1`, then
`X^2=Y-1\implies X=√(Y-1)`, where we take the positive root because we know
`X` can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that
`Y` can take on the values
`1^2+1=2`,
`2^2+1=5`, and
`5^2+1=26`. At these values of
`Y`, we would have the same probability as we did for the respective value of
`X`. That is,


`\mathbb P(Y=y)=\begin{cases}\frac c3&\text{for }y=2\\\\\frac c6&\text{for }y=5\\\\\frac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}`

Part (5) is incomplete, so I'll stop here.