Question

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Alice lost her phone a wek ago. When she finally got a new phone with a replaced SIM card, she found she got a thousand new messages, many of whichase Just spum She wanted fiber out the乎am. To her sadness, she lost the contacts as wel and could not know which messages ae from her friends. Aice went through 12 messages and moted down 4 observations for each message in Table 1) Y: whether the message is spam: I for yes, 0 for no, (2) H whether the sender's number is local. I for yes. O fof no: 3) U:whether the message has an URL link: I for yes, 0 for no, 4) L: whether the message is long: I for having more than 40 characters, O otherwise Table I observations of the messages Y H UL 0.66 A. [18 points] Complcte Table 2 with probabilkies using the information present in Table 1. The conditional 1Y0.667 and the pricr probability P(Y-0)- 0.5 has been filled out for probabilty P(H B. |7 points] Alice wants to determine if a message is spam based on its characteristics: whether or not the sender's number is local, the message has an URL link, or the message is long. In terms of probabilistic reasoning. we can formulate the question as evaluating the condaional probabiity PIYH.U.L). We say that the message is a spam if F"(Y = 1 I H. U. L) > P(Y 0' H. U. L). Suppose we have a message from a non-kcal number 《H = 0). which has an URL link(U 1 ) andis not long (L-. We wantb determine if te message is a span. Compute P(Y = 1 I H = 0. U 1-1 0) and F"(Y-0 I H 0, U 1. L 0) using the 12 examples in Table l would you conclude that the message is a spam? C. 15 points] Now, ignore Table 1, and consider all possibilities you can fi in Table 2 Are there any cons๕ants on the values that we must respect when assigning these values? In other words, cai we fill in Table 2 with 8 arbitrary real values? If not, are there any constraints on some values that we must folow?

Answer 1

$A.~P(H=1|Y=1)=\frac{4}{6}=0.667,~P(U=1|Y=1)=\frac{5}{6}=0.833,\\~P(L=1|Y=1)=\frac{2}{6}=0.333,~P(Y=1)=6/12=0.5\\ P(H=1|Y=0)=\frac{2}{6}=0.333,~P(U=1|Y=0)=\frac{3}{6}=0.5,\\~P(L=1|Y=0)=\frac{2}{6}=0.333,~P(Y=0)=6/12=0.5\\ B.~P(Y=1|H=0,U=1,L=0)=\frac{1}{2}\\ P(Y=0|H=0,U=1,L=0)=\frac{1}{2}\\ Since~P(Y=1|H=0,U=1,L=0)=P(Y=0|H=0,U=1,L=0)~\\ so~the~message~is~not~a~spam.\\ C.~From~Table~2~it~is~observed~that\\ P(U=1|Y=1)>P(H=1|Y=1)>P(L=1|Y=1)~i.e.\\ Given~that~message~is~spam,~probability~that~the~message~has~an~URL~is\\ biggest~relative~to~other~two.~However\\ Given~that~message~is~spam,~probability~that~the~message~is~long~is\\relatively~small~than~other~two.\\ So~we~should~respect~these~two~points.$