Question

2 .Define the probability density functions (pdfs) of the independent random variables X and Y as:

f_X(x)={ (1/5).(1-(x/10)), 0≤x≤10 f_{Y(y)={ (1/5).(y/10) ,} 0≤y≤10

0 , otherwise for both f_X(x) and f_Y(y)

(a) Suppose that you are playing a game where you want to maximize your number of points , and you can have your score be X or Y . Which would you choose?

(b) Suppose I am confused in part (a), so I flip an unfair coin, with P (“heads”) = 3/4 , P (“tails”) = 1/4 , to select which probability density function to draw my score from: if the coin is “heads”, I choose X; if the coin is “tails”, I choose Y . What is the probability that I score more than 5 points.

(c) Suppose generate both X and Y . What is the probability that Y is bigger than X? That is, what  is P(Y >X)?

Answer 1

a) PDFs for X is { (1/5).(1-(x/10)) and for Y is { (1/5).(y/10). Both are appliacble for 0≤x≤10 and 0≤y≤10.

As we can evaluate

For X = 0

P(X=0) = { (1/5).(1-(0/10)) = 0.2 and

For X = 10

P(X=0) = { (1/5).(1-(10/10)) = 0

again,

For Y = 0

P(Y=0) =  { (1/5).(0/10) = 0

For Y = 10

P(Y=0) =  { (1/5).(10/10) = 0.2

As we increase the value of the random variable, PDF for X gives successively lower value of the probability and PDF for Y gives successively higher values of the probability.

Hence, we can say that PDF for X is a decreasing function and PDF for Y is an increasing function.

Now, for X = 5 and Y = 5

P(X=5) = { (1/5).(1-(5/10)) = 0.1

P(Y=5) = {1/5).(5/10} = 0.1

PDF for X is decreasing and Y is increasing. Both are defined from 0 to 10 and both have the same value at 5.

Hence, we can say that for values higher than 5, the probability of achieving it through the random variable Y is more than through the random variable X. And since we want to maximize the points, random variable Y would be a better option to choose.

b) The probability of scoring more than 5 points would be the cumulative probability of achieving it through the random variable X and Y, combined with the probability of getting X or Y as the preferred variable with the help of the toss.

Probability of scoring more than 5 points through X

=

=

=

=

Probability of scoring more than 5 points through Y

Hence, the total probability of scoring more than 5 points = 0.2 + 0.15 = 0.35 (answer)

c) P(Y>X) has to be calculated for all possible values of X and the corresponding values of Y for each X.

For X = 0

P(Y>X) = P(Y>0)P(X = 0)

For X = 1

P(Y>X) = P(Y>1) P(X=1)

.

.

.

For X = 9

P(Y>X) = P(Y>9) P(X=9)

and For X = 10

P(Y>X) = P(Y>10) = 0

Adding up all these probabilities for Y will give us the required probability

for x = 0

for x = 1

for x = 2

for x = 3

for x = 4

for x = 5

for x = 6

for x = 7

for x = 8

for x = 9

for x =10 , p(y>10).p(x=10) = 0 as the maximum possible value of Y = 10

Adding all the probabilities will give the required probability P(Y>X) = 1.0119

Now, here is a problem Probability can't be greater than 1

This is happening because the PDF for Y is not correct or possible.

Since Y can take values between 0 to 10 hence,

But on the calculation, we get   which is not correct.

The above process is right. The PDF for Y is wrongly given in the question.

Thank You!

PLease Upvote.

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