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Let X be a continuous rv with PDF $$Fx(X) =\begin{cases}4x^3 & 0<x≤1 \\ 0 & \text{otherwise}.\end{cases}$$After many tries, I've come to the conclusion that the answer is zero since the question is asking us to find a probability that is outside the given interval. Anyone have any other idea or is my answer right?

gt6989b
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Bee
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2 Answers2

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Your PDF has support only on $[0,1]$ so $X$ lies between $0$ and $1$ with probability one: $$ P(0\le X\le 1) = 1.$$ This means in particular that $$ P(X>13) = 0$$ (if it's between zero and one it's definitely not greater than thirteen) and $$ P(X\le 23)=1$$ (if it's between zero and one it's definitely less than or equal to twenty three).

If you tried to apply the naive formula you'd get $$P(X\le 23\mid X>13) = \frac{P(X\le 23)}{P(X>13)} = \frac{1}{0}$$ so that's a problem. You're trying to take a conditional probability with respect to the event $X>13,$ but that event has probability zero. This is why in the fine print of the definition of conditional probability with respect to an event, it stipulates that the event you're conditioning on must have nonzero probability. Otherwise the conditional probability is undefined.

spaceisdarkgreen
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I think you really mean $f_X(x)$, not $F_X(x)$, and the proof is that $$ \int_0^1 4x^3dx = 1. $$

Now that figured out, note that by definition, $$\mathbb{P}[X \le 23] = F(23)$$ and $$\mathbb{P}[X>13] = 1 - F(13),$$ where $$ F_X(x) = \int_{-\infty}^x f_X(s)ds. $$

Can you

  1. finish the integral to get an expression for $F_X(x)$;
  2. plug your event into the definition of conditional probability and evaluate the probabilities in the numerator and denominator of the resulting expression;
  3. simplify the result to find the answer?
gt6989b
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