Probability calculations verification request: a distribution function related problem
Multi tool use
$begingroup$
Let $X$ have distribution function $F(x)$ expressed by
begin{cases}
0 & mbox{if } x < 0 \
x/2 & mbox{if } 0leq x leq 2 \
1 & mbox{if } x geq 2
end{cases}
and let $Y = X^{2}$. Find
(a) $textbf{P}(frac{1}{2}leq Xleq frac{3}{2})$
(b) $textbf{P}(1leq X < 2)$
(c) $textbf{P}(Yleq X)$
(d) $textbf{P}(Xleq 2Y)$
(e) $textbf{P}(X+Yleqfrac{3}{4})$
(f) The distribution of $Z = sqrt{X}$.
MY ATTEMPT
(a) $displaystyletextbf{P}(0.5leq Xleq 1.5) = F(1.5) - lim_{xuparrow 0.5} F(x) = 0.75 - 0.25 = 0.5$
(b) $displaystyletextbf{P}(1leq X < 2) = lim_{xuparrow 2}F(x) - lim_{xuparrow 1}F(x) = 1 - 0.5 = 0.5$
(c) $displaystyletextbf{P}(Yleq X) = textbf{P}(X^{2} leq X) = textbf{P}(X(X-1) leq 0) = textbf{P}(0 leq Xleq 1)$
$displaystyletherefore textbf{P}(Yleq X) = F(1) - lim_{xuparrow 0}F(x) = 0.5 - 0 = 0.5$
(d) $displaystyletextbf{P}(X leq 2Y) = textbf{P}(Xleq 2X^{2}) = textbf{P}(X(2X-1)geq0) = 1 - textbf{P}(X<0.5)$
$displaystyletherefore textbf{P}(Xleq 2Y) = 1 - lim_{xuparrow 0.5}F(x) = 1 - 0 .25 = 0.75$
(e) $textbf{P}(X+Yleqfrac{3}{4}) = textbf{P}(4X^{2} + 4X - 3leq 0) = textbf{P}(-1.5 leq Xleq 0.5)$
$displaystyletherefore textbf{P}(X+Yleq 0.75) = F(0.5) - lim_{xuparrow-1.5}F(x) = 0.25 - 0 = 0.25$
(f) I am not sure how to solve it. Here is my attempt
begin{align*}
textbf{P}(Zleq z) = textbf{P}(sqrt{X}leq z) = textbf{P}(Xleq z^{2}) = F_{X}(z^{2})
end{align*}
More precisely, the distribution function of $Z$ is given by
begin{cases}
0 & mbox{if } z < 0 \
z^{2}/2 & mbox{if } 0leq z leq sqrt{2} \
1 & mbox{if } z geqsqrt{2}
end{cases}
Could someone please double-check my results? I thank in advance for any contribution.
EDIT
I have edited the answer according to Kavi's contribution. Any other contribution is equally welcome as well.
probability probability-theory proof-verification probability-distributions
$endgroup$
add a comment |
$begingroup$
Let $X$ have distribution function $F(x)$ expressed by
begin{cases}
0 & mbox{if } x < 0 \
x/2 & mbox{if } 0leq x leq 2 \
1 & mbox{if } x geq 2
end{cases}
and let $Y = X^{2}$. Find
(a) $textbf{P}(frac{1}{2}leq Xleq frac{3}{2})$
(b) $textbf{P}(1leq X < 2)$
(c) $textbf{P}(Yleq X)$
(d) $textbf{P}(Xleq 2Y)$
(e) $textbf{P}(X+Yleqfrac{3}{4})$
(f) The distribution of $Z = sqrt{X}$.
MY ATTEMPT
(a) $displaystyletextbf{P}(0.5leq Xleq 1.5) = F(1.5) - lim_{xuparrow 0.5} F(x) = 0.75 - 0.25 = 0.5$
(b) $displaystyletextbf{P}(1leq X < 2) = lim_{xuparrow 2}F(x) - lim_{xuparrow 1}F(x) = 1 - 0.5 = 0.5$
(c) $displaystyletextbf{P}(Yleq X) = textbf{P}(X^{2} leq X) = textbf{P}(X(X-1) leq 0) = textbf{P}(0 leq Xleq 1)$
$displaystyletherefore textbf{P}(Yleq X) = F(1) - lim_{xuparrow 0}F(x) = 0.5 - 0 = 0.5$
(d) $displaystyletextbf{P}(X leq 2Y) = textbf{P}(Xleq 2X^{2}) = textbf{P}(X(2X-1)geq0) = 1 - textbf{P}(X<0.5)$
$displaystyletherefore textbf{P}(Xleq 2Y) = 1 - lim_{xuparrow 0.5}F(x) = 1 - 0 .25 = 0.75$
(e) $textbf{P}(X+Yleqfrac{3}{4}) = textbf{P}(4X^{2} + 4X - 3leq 0) = textbf{P}(-1.5 leq Xleq 0.5)$
$displaystyletherefore textbf{P}(X+Yleq 0.75) = F(0.5) - lim_{xuparrow-1.5}F(x) = 0.25 - 0 = 0.25$
(f) I am not sure how to solve it. Here is my attempt
begin{align*}
textbf{P}(Zleq z) = textbf{P}(sqrt{X}leq z) = textbf{P}(Xleq z^{2}) = F_{X}(z^{2})
end{align*}
More precisely, the distribution function of $Z$ is given by
begin{cases}
0 & mbox{if } z < 0 \
z^{2}/2 & mbox{if } 0leq z leq sqrt{2} \
1 & mbox{if } z geqsqrt{2}
end{cases}
Could someone please double-check my results? I thank in advance for any contribution.
EDIT
I have edited the answer according to Kavi's contribution. Any other contribution is equally welcome as well.
probability probability-theory proof-verification probability-distributions
$endgroup$
add a comment |
$begingroup$
Let $X$ have distribution function $F(x)$ expressed by
begin{cases}
0 & mbox{if } x < 0 \
x/2 & mbox{if } 0leq x leq 2 \
1 & mbox{if } x geq 2
end{cases}
and let $Y = X^{2}$. Find
(a) $textbf{P}(frac{1}{2}leq Xleq frac{3}{2})$
(b) $textbf{P}(1leq X < 2)$
(c) $textbf{P}(Yleq X)$
(d) $textbf{P}(Xleq 2Y)$
(e) $textbf{P}(X+Yleqfrac{3}{4})$
(f) The distribution of $Z = sqrt{X}$.
MY ATTEMPT
(a) $displaystyletextbf{P}(0.5leq Xleq 1.5) = F(1.5) - lim_{xuparrow 0.5} F(x) = 0.75 - 0.25 = 0.5$
(b) $displaystyletextbf{P}(1leq X < 2) = lim_{xuparrow 2}F(x) - lim_{xuparrow 1}F(x) = 1 - 0.5 = 0.5$
(c) $displaystyletextbf{P}(Yleq X) = textbf{P}(X^{2} leq X) = textbf{P}(X(X-1) leq 0) = textbf{P}(0 leq Xleq 1)$
$displaystyletherefore textbf{P}(Yleq X) = F(1) - lim_{xuparrow 0}F(x) = 0.5 - 0 = 0.5$
(d) $displaystyletextbf{P}(X leq 2Y) = textbf{P}(Xleq 2X^{2}) = textbf{P}(X(2X-1)geq0) = 1 - textbf{P}(X<0.5)$
$displaystyletherefore textbf{P}(Xleq 2Y) = 1 - lim_{xuparrow 0.5}F(x) = 1 - 0 .25 = 0.75$
(e) $textbf{P}(X+Yleqfrac{3}{4}) = textbf{P}(4X^{2} + 4X - 3leq 0) = textbf{P}(-1.5 leq Xleq 0.5)$
$displaystyletherefore textbf{P}(X+Yleq 0.75) = F(0.5) - lim_{xuparrow-1.5}F(x) = 0.25 - 0 = 0.25$
(f) I am not sure how to solve it. Here is my attempt
begin{align*}
textbf{P}(Zleq z) = textbf{P}(sqrt{X}leq z) = textbf{P}(Xleq z^{2}) = F_{X}(z^{2})
end{align*}
More precisely, the distribution function of $Z$ is given by
begin{cases}
0 & mbox{if } z < 0 \
z^{2}/2 & mbox{if } 0leq z leq sqrt{2} \
1 & mbox{if } z geqsqrt{2}
end{cases}
Could someone please double-check my results? I thank in advance for any contribution.
EDIT
I have edited the answer according to Kavi's contribution. Any other contribution is equally welcome as well.
probability probability-theory proof-verification probability-distributions
$endgroup$
Let $X$ have distribution function $F(x)$ expressed by
begin{cases}
0 & mbox{if } x < 0 \
x/2 & mbox{if } 0leq x leq 2 \
1 & mbox{if } x geq 2
end{cases}
and let $Y = X^{2}$. Find
(a) $textbf{P}(frac{1}{2}leq Xleq frac{3}{2})$
(b) $textbf{P}(1leq X < 2)$
(c) $textbf{P}(Yleq X)$
(d) $textbf{P}(Xleq 2Y)$
(e) $textbf{P}(X+Yleqfrac{3}{4})$
(f) The distribution of $Z = sqrt{X}$.
MY ATTEMPT
(a) $displaystyletextbf{P}(0.5leq Xleq 1.5) = F(1.5) - lim_{xuparrow 0.5} F(x) = 0.75 - 0.25 = 0.5$
(b) $displaystyletextbf{P}(1leq X < 2) = lim_{xuparrow 2}F(x) - lim_{xuparrow 1}F(x) = 1 - 0.5 = 0.5$
(c) $displaystyletextbf{P}(Yleq X) = textbf{P}(X^{2} leq X) = textbf{P}(X(X-1) leq 0) = textbf{P}(0 leq Xleq 1)$
$displaystyletherefore textbf{P}(Yleq X) = F(1) - lim_{xuparrow 0}F(x) = 0.5 - 0 = 0.5$
(d) $displaystyletextbf{P}(X leq 2Y) = textbf{P}(Xleq 2X^{2}) = textbf{P}(X(2X-1)geq0) = 1 - textbf{P}(X<0.5)$
$displaystyletherefore textbf{P}(Xleq 2Y) = 1 - lim_{xuparrow 0.5}F(x) = 1 - 0 .25 = 0.75$
(e) $textbf{P}(X+Yleqfrac{3}{4}) = textbf{P}(4X^{2} + 4X - 3leq 0) = textbf{P}(-1.5 leq Xleq 0.5)$
$displaystyletherefore textbf{P}(X+Yleq 0.75) = F(0.5) - lim_{xuparrow-1.5}F(x) = 0.25 - 0 = 0.25$
(f) I am not sure how to solve it. Here is my attempt
begin{align*}
textbf{P}(Zleq z) = textbf{P}(sqrt{X}leq z) = textbf{P}(Xleq z^{2}) = F_{X}(z^{2})
end{align*}
More precisely, the distribution function of $Z$ is given by
begin{cases}
0 & mbox{if } z < 0 \
z^{2}/2 & mbox{if } 0leq z leq sqrt{2} \
1 & mbox{if } z geqsqrt{2}
end{cases}
Could someone please double-check my results? I thank in advance for any contribution.
EDIT
I have edited the answer according to Kavi's contribution. Any other contribution is equally welcome as well.
probability probability-theory proof-verification probability-distributions
probability probability-theory proof-verification probability-distributions
edited Jan 22 at 19:52
user1337
asked Jan 21 at 23:14
user1337user1337
46110
46110
add a comment |
add a comment |
1 Answer
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$begingroup$
I think there is a mistake in e). $4X^{2}+4X-3 leq 0$ gives $-frac 3 2 leq X leq frac 1 2$. In f) you should replace the condition $0 leq z leq 2$ by $0 leq z^{2} leq 2$ or $0 leq z leq sqrt 2$.
$endgroup$
$begingroup$
Thanks, Kavi, for the contribution. I have edited the question according to your observations.
$endgroup$
– user1337
Jan 21 at 23:51
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think there is a mistake in e). $4X^{2}+4X-3 leq 0$ gives $-frac 3 2 leq X leq frac 1 2$. In f) you should replace the condition $0 leq z leq 2$ by $0 leq z^{2} leq 2$ or $0 leq z leq sqrt 2$.
$endgroup$
$begingroup$
Thanks, Kavi, for the contribution. I have edited the question according to your observations.
$endgroup$
– user1337
Jan 21 at 23:51
add a comment |
$begingroup$
I think there is a mistake in e). $4X^{2}+4X-3 leq 0$ gives $-frac 3 2 leq X leq frac 1 2$. In f) you should replace the condition $0 leq z leq 2$ by $0 leq z^{2} leq 2$ or $0 leq z leq sqrt 2$.
$endgroup$
$begingroup$
Thanks, Kavi, for the contribution. I have edited the question according to your observations.
$endgroup$
– user1337
Jan 21 at 23:51
add a comment |
$begingroup$
I think there is a mistake in e). $4X^{2}+4X-3 leq 0$ gives $-frac 3 2 leq X leq frac 1 2$. In f) you should replace the condition $0 leq z leq 2$ by $0 leq z^{2} leq 2$ or $0 leq z leq sqrt 2$.
$endgroup$
I think there is a mistake in e). $4X^{2}+4X-3 leq 0$ gives $-frac 3 2 leq X leq frac 1 2$. In f) you should replace the condition $0 leq z leq 2$ by $0 leq z^{2} leq 2$ or $0 leq z leq sqrt 2$.
answered Jan 21 at 23:40
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
62.7k42262
$begingroup$
Thanks, Kavi, for the contribution. I have edited the question according to your observations.
$endgroup$
– user1337
Jan 21 at 23:51
add a comment |
$begingroup$
Thanks, Kavi, for the contribution. I have edited the question according to your observations.
$endgroup$
– user1337
Jan 21 at 23:51
$begingroup$
Thanks, Kavi, for the contribution. I have edited the question according to your observations.
$endgroup$
– user1337
Jan 21 at 23:51
$begingroup$
Thanks, Kavi, for the contribution. I have edited the question according to your observations.
$endgroup$
– user1337
Jan 21 at 23:51
add a comment |
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