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Suppose that a random variable X can have each of the seven values −3, −2, −1, 0, 1, 2, 3 with equal probability. Determine the p f. Of \(Y = {X^2} - X\)

Short Answer

Expert verified

The p f of \(Y = {X^2} - X = \frac{1}{7}\)

Step by step solution

01

given information

Random variable X can have each seven values -3,-2,-1,0,1,2,3

02

Compute the probability

Possible values for Y are 0, 2, 6,12.

\(\begin{aligned}{\rm P}\left( {Y = 0} \right) &= {\rm P}\left( {X = 1{\rm{ }}or{\rm{ }}X = 0} \right)\\ &= \frac{2}{7}\end{aligned}\)

\(\begin{aligned}{\rm P}\left( {Y = 2} \right) &= {\rm P}\left( {X = - 1{\rm{ }}or{\rm{ }}X = 2} \right)\\ &= \frac{2}{7}\end{aligned}\)

\(\begin{aligned}{\rm P}\left( {Y = 6} \right) &= {\rm P}\left( {X = - 2{\rm{ }}or{\rm{ }}X = 3} \right)\\ &= \frac{2}{7}\end{aligned}\)

\(\begin{aligned}{\rm P}\left( {Y = 12} \right) &= {\rm P}\left( {{\rm{ }}X = - 3} \right)\\ &= \frac{1}{7}\end{aligned}\)

Therefore the p f of\(Y = {X^2} - X = \frac{1}{7}\)

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Most popular questions from this chapter

Suppose that a Markov chain has four states 1, 2, 3, 4, and stationary transition probabilities as specified by the following transition matrix

\(p = \left[ {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{4}}&0&{\frac{1}{2}}\\0&1&0&0\\{\frac{1}{2}}&0&{\frac{1}{2}}&0\\{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\end{array}} \right]\):

a.If the chain is in state 3 at a given timen, what is the probability that it will be in state 2 at timen+2?

b.If the chain is in state 1 at a given timen, what is the probability it will be in state 3 at timen+3?

Suppose that a point (X, Y) is chosen at random from the disk S defined as follows:

\(S = \left\{ {\left( {x,y} \right) :{{\left( {x - 1} \right)}^2} + {{\left( {y + 2} \right)}^2} \le 9} \right\}.\) Determine (a) the conditional pdf of Y for every given value of X, and (b) \({\rm P}\left( {Y > 0|x = 2} \right)\)

Suppose that a random variableXhas the uniform distributionon the integers 10, . . . ,20. Find the probability thatXis even.

The definition of the conditional p.d.f. of X given\({\bf{Y = y}}\)is arbitrary if\({{\bf{f}}_{\bf{2}}}\left( {\bf{y}} \right){\bf{ = 0}}\). The reason that this causes no serious problem is that it is highly unlikely that we will observe Y close to a value\({{\bf{y}}_{\bf{0}}}\)such that\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{0}}}} \right){\bf{ = 0}}\). To be more precise, let\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{0}}}} \right){\bf{ = 0}}\), and let\({{\bf{A}}_{\bf{0}}}{\bf{ = }}\left( {{{\bf{y}}_{\bf{0}}}{\bf{ - }} \in {\bf{,}}{{\bf{y}}_{\bf{0}}}{\bf{ + }} \in } \right)\). Also, let\({{\bf{y}}_{\bf{1}}}\)be such that\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{1}}}} \right){\bf{ > 0}}\), and let\({{\bf{A}}_{\bf{1}}}{\bf{ = }}\left( {{{\bf{y}}_{\bf{1}}}{\bf{ - }} \in {\bf{,}}{{\bf{y}}_{\bf{1}}}{\bf{ + }} \in } \right)\). Assume that\({{\bf{f}}_{\bf{2}}}\)is continuous at both\({{\bf{y}}_{\bf{0}}}\)and\({{\bf{y}}_{\bf{1}}}\).

Show that

\(\mathop {{\bf{lim}}}\limits_{ \in \to {\bf{0}}} \,\frac{{{\bf{Pr}}\left( {{\bf{Y}} \in {{\bf{A}}_{\bf{0}}}} \right)}}{{{\bf{Pr}}\left( {{\bf{Y}} \in {{\bf{A}}_{\bf{1}}}} \right)}}{\bf{ = 0}}{\bf{.}}\)

That is, the probability that Y is close to\({{\bf{y}}_{\bf{0}}}\)is much smaller than the probability that Y is close to\({{\bf{y}}_{\bf{1}}}\).

Suppose that a random variableXhas the binomial distribution with parametersn=15 andp=0.5. Find Pr(X <6).

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