P(x | Y=1) and P(x | Y=2) is the conditional probability of X given Y has occurred. To calculate expected value of X –
1st Case (Y=1) = (0.2 x 0.3 x 0) + (0.4 x 0.3 x 5) + (0.4 x 0.3 x 10) = 1.8
Which means: The probability of (Y = 1) occurring is 0.3 or 30%. If Y occurs, probability of X being 0 is 0.2 So, joint probability X=0 is 0.2 x 0.3 x 0. (1st bracket above) and same for the 2nd and 3rd bracket.
Similarly,
2nd Case (Y=2) = (0.1 x 0.7 x 0) + (0.8 x 0.7 x 5) + (0.1 x 0.7 x 10) = 3.5
Expected Value of X = 1.8 + 3.5 = 5.3
This is my understanding, please confirm me whether the answer is correct 🙂
Hi Saideep
P(x | Y=1) and P(x | Y=2) is the conditional probability of X given Y has occurred. To calculate expected value of X –
1st Case (Y=1) = (0.2 x 0.3 x 0) + (0.4 x 0.3 x 5) + (0.4 x 0.3 x 10) = 1.8
Which means: The probability of (Y = 1) occurring is 0.3 or 30%. If Y occurs, probability of X being 0 is 0.2 So, joint probability X=0 is 0.2 x 0.3 x 0. (1st bracket above) and same for the 2nd and 3rd bracket.
Similarly,
2nd Case (Y=2) = (0.1 x 0.7 x 0) + (0.8 x 0.7 x 5) + (0.1 x 0.7 x 10) = 3.5
Expected Value of X = 1.8 + 3.5 = 5.3
This is my understanding, please confirm me whether the answer is correct 🙂
Yes, answer is correct.
Thank you.