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Textbook Question: \begin{equation} f(x,y) = \begin{cases} x + y & \text{if $0\le x \le1$, $0\le y \le1$ } \\ 0 & \text{otherwise} \end{cases} \end{equation}

What is $P(X < \frac{1}{4} \mid Y = \frac{1}{3})$?

[edit: added solution for more clarity]

I'm CAN SOLVE the question by applying $P(X \in A \mid Y=y) = \int_A f(_X\mid_Y) (x\mid y)dx$

Deriving (1) $$ (1) - f(_X\mid_Y) (x\mid y) = \frac{f(_X,_Y)(x,y)}{f_Y(y)} = \frac{x+y}{y+\frac{1}{2}} $$ $$ P(X < \frac{1}{4} \mid Y=\frac{1}{3}) = \int_0^\frac{1}{4} f(_X\mid_Y) (x\mid \frac{1}{3})dx $$ $$ = \int_0^\frac{1}{4}\frac{x+\frac{1}{3}}{\frac{1}{3}+\frac{1}{2}}dx$$ $$ = \frac{11}{80} $$

My question is Conceptual However:

  1. if Y is a Continuous Random Variable $P(Y=y) = 0$; meaning $P(Y=\frac{1}{3}) = 0$. Hence Event $Y = \frac{1}{3}$ CANNOT occur.
  2. hence in the above Question I can't conceptually appreciate how to condition on an Event i.e $Y = \frac{1}{3}$ that CANNOT occur.

Thanks in advance for the explanations.

p/s: I had read previous Questions: Conditional probability when conditioning on continuous-discrete variables & Conditional probability combining discrete and continuous random variables But couldn't figure a clear answer hence raise this Q.

Apologies if it's too noob a question; just starting Statistics.

seven
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  • Hi! I don't wish to downvote as you are new user! But to get answers here, we require some sort of understanding or info from your side. Like what have you understood from this problem. – Elizabeth Huffman Mar 29 '23 at 14:48
  • thank you for the comment @ElizabethHuffman. I've updated the Question for more clarity; The ask was not the answer to the question. That I have done. My Question is conceptually comprehending the use of a discrete value on a continuous variable please. – seven Mar 29 '23 at 15:53
  • Now its perfect! – Elizabeth Huffman Mar 29 '23 at 16:08
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    Probability 0 events can occur. From a frequentist perspective, they just occur at a rate of $o(n)$, where $n$ is the number of times you repeat the experiment. You are right that you cannot condition on a zero probability event. In fact, the question you were given is not well-defined. However, the mathematical resolution is rather complicated (see regular conditional distributions). Conditional probabilities are therefore ignored for the most part in more elementary probability texts. – Andrew Mar 29 '23 at 16:20
  • thank you @AndrewZhang. That was helpful. – seven Mar 29 '23 at 16:33

1 Answers1

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For completeness answering my Question based on the comments provided.

if Y is a Continuous Random Variable $P(Y=y)=0$; meaning $P(Y=\frac{1}{3})=0$. Hence Event $Y=\frac{1}{3}$ CANNOT occur.

Probability 0 events can occur. From a frequentist perspective, they just occur at a rate of () o(n), where n is the number of times you repeat the experiment.

hence in the above Question I can't conceptually appreciate how to condition on an Event i.e $Y=\frac{1}{3}$ that CANNOT occur.

assuming $P(Y=\frac{1}{3}) > 0$ we can use the (∈∣=)=∫(∣)(∣). Given the Mathematical resolution of this is complicated better taken as a definition for entry level.

seven
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