Cumulative Distribution Function and

Question

The demand, $X$, for a firm’s product is a random variable with density $f(x) = 2x$ for $0 ≤ x ≤ 1$. The corresponding cumulative distribution function is $F (x) = x^2$ for $0 ≤ x ≤ 1$. The firm’s profits, $Y$ , as a function of $X$ are $Y = \sqrt X − \frac14$ . What is the probability that the firm realizes a loss?

I see that the derivative of the cdf is the density, but I'm not sure how to apply this information to the profit equation. I'm guessing that I need to set the profit function to zero.

$P(X<1/16)=F(1/16)=1/256 – Alexander Vigodner Jun 25 '14 at 06:13 — Alexander Vigodner, Jun 25 '14 at 06:13

score 0 · Answer 1 · answered Jun 25 '14 at 06:13

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Well, the probability of loss can be said as $P(Y \leq 0)$, which is $P(\sqrt{X}-1/4 \leq 0)=P(X \leq 1/16)=F(1/16)=1/256.$ Correct me if I'm wrong!

answered Jun 25 '14 at 06:13

puru

1,109

How did you get the 1/16? I follow after that part. Thanks! This is very new to me. – Rosie E Jun 25 '14 at 06:18
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Since $\sqrt{X}\leq 1/4 \implies X \leq 1/16$. I hope this helps. – puru Jun 25 '14 at 06:50

Cumulative Distribution Function and

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