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CDF from PDF of a function
Finding a CDF given a PDFProducing a CDF from a given PDFIs this a joint distribution? Trying pdf and cdfHow obtain a CDF from a PDF?How to find the CDF and PDFStruggling with this PDF and finding the CDFFind the CDF given the PDF (with indicator function)Finding the cdf from pdfCDF and PDF of standard normal random variableCalculating cdf from a piecewise pdf
$begingroup$
I have this problem where I need to graph the CDF, for that I need to find the constant $c$. The formula below is a PDF:
$f(x) = c(x^2+1)spacespacespace ifspace X in [0,1];space$ otherwise $0$
My attempt:
$P(0leq X leq 1) = int_0^1c*(x^2+1) = 1 Rightarrow cx+frac cx^33 = 1 Rightarrow c = 3/4$
The problem here is that when we put $x=1$ we get $f(1) = frac 32 > 1$, which is wrong as probability cannot be higher than $1$. I need to construct the CDF, but because of that, I cannot do it.
probability probability-distributions
edited Mar 21 at 20:34
GNUSupporter 8964民主女神 地下教會
14k82651
asked Mar 21 at 20:07
Tigran MinasyanTigran Minasyan
228
$endgroup$
add a comment |
$begingroup$
I have this problem where I need to graph the CDF, for that I need to find the constant $c$. The formula below is a PDF:
$f(x) = c(x^2+1)spacespacespace ifspace X in [0,1];space$ otherwise $0$
My attempt:
$P(0leq X leq 1) = int_0^1c*(x^2+1) = 1 Rightarrow cx+frac cx^33 = 1 Rightarrow c = 3/4$
The problem here is that when we put $x=1$ we get $f(1) = frac 32 > 1$, which is wrong as probability cannot be higher than $1$. I need to construct the CDF, but because of that, I cannot do it.
probability probability-distributions
edited Mar 21 at 20:34
GNUSupporter 8964民主女神 地下教會
14k82651
asked Mar 21 at 20:07
Tigran MinasyanTigran Minasyan
228
$endgroup$
add a comment |
$begingroup$
I have this problem where I need to graph the CDF, for that I need to find the constant $c$. The formula below is a PDF:
$f(x) = c(x^2+1)spacespacespace ifspace X in [0,1];space$ otherwise $0$
My attempt:
$P(0leq X leq 1) = int_0^1c*(x^2+1) = 1 Rightarrow cx+frac cx^33 = 1 Rightarrow c = 3/4$
The problem here is that when we put $x=1$ we get $f(1) = frac 32 > 1$, which is wrong as probability cannot be higher than $1$. I need to construct the CDF, but because of that, I cannot do it.
probability probability-distributions
edited Mar 21 at 20:34
GNUSupporter 8964民主女神 地下教會
14k82651
asked Mar 21 at 20:07
Tigran MinasyanTigran Minasyan
228
$endgroup$
I have this problem where I need to graph the CDF, for that I need to find the constant $c$. The formula below is a PDF:
$f(x) = c(x^2+1)spacespacespace ifspace X in [0,1];space$ otherwise $0$
My attempt:
$P(0leq X leq 1) = int_0^1c*(x^2+1) = 1 Rightarrow cx+frac cx^33 = 1 Rightarrow c = 3/4$
The problem here is that when we put $x=1$ we get $f(1) = frac 32 > 1$, which is wrong as probability cannot be higher than $1$. I need to construct the CDF, but because of that, I cannot do it.
probability probability-distributions
probability probability-distributions
edited Mar 21 at 20:34
GNUSupporter 8964民主女神 地下教會
14k82651
asked Mar 21 at 20:07
Tigran MinasyanTigran Minasyan
228
edited Mar 21 at 20:34
GNUSupporter 8964民主女神 地下教會
14k82651
asked Mar 21 at 20:07
Tigran MinasyanTigran Minasyan
228
edited Mar 21 at 20:34
GNUSupporter 8964民主女神 地下教會
14k82651
edited Mar 21 at 20:34
GNUSupporter 8964民主女神 地下教會
14k82651
edited Mar 21 at 20:34
GNUSupporter 8964民主女神 地下教會
14k82651
14k82651
asked Mar 21 at 20:07
Tigran MinasyanTigran Minasyan
228
asked Mar 21 at 20:07
Tigran MinasyanTigran Minasyan
228
asked Mar 21 at 20:07
Tigran MinasyanTigran Minasyan
228
228
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The CDF cannot exceed $1$, but the PDF can. Your CDF is $frac34x+frac14x^3$ on $[0,,1]$, just as you calculated.
answered Mar 21 at 20:11
J.G.J.G.
32.6k23250
$endgroup$
$begingroup$
Then how can you find the CDF from the PDF?
$endgroup$
– Tigran Minasyan
Mar 21 at 20:12
$begingroup$
@TigranMinasyan You basically did it already; see my edit.
$endgroup$
– J.G.
Mar 21 at 20:13
1
$begingroup$
Indeed. The CDF is $F(x)~=mathbf 1_0leq xlt 1~int_0^x c~(s^2+1)~mathrm d s+mathbf 1_1leq x\= tfrac 34(tfrac 13x^3+x)~mathbf 1_0leq xlt 1+mathbf 1_1leq x$
$endgroup$
– Graham Kemp
Mar 21 at 21:47
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
The CDF cannot exceed $1$, but the PDF can. Your CDF is $frac34x+frac14x^3$ on $[0,,1]$, just as you calculated.
answered Mar 21 at 20:11
J.G.J.G.
32.6k23250
$endgroup$
$begingroup$
Then how can you find the CDF from the PDF?
$endgroup$
– Tigran Minasyan
Mar 21 at 20:12
$begingroup$
@TigranMinasyan You basically did it already; see my edit.
$endgroup$
– J.G.
Mar 21 at 20:13
1
$begingroup$
Indeed. The CDF is $F(x)~=mathbf 1_0leq xlt 1~int_0^x c~(s^2+1)~mathrm d s+mathbf 1_1leq x\= tfrac 34(tfrac 13x^3+x)~mathbf 1_0leq xlt 1+mathbf 1_1leq x$
$endgroup$
– Graham Kemp
Mar 21 at 21:47
add a comment |
$begingroup$
The CDF cannot exceed $1$, but the PDF can. Your CDF is $frac34x+frac14x^3$ on $[0,,1]$, just as you calculated.
answered Mar 21 at 20:11
J.G.J.G.
32.6k23250
$endgroup$
$begingroup$
Then how can you find the CDF from the PDF?
$endgroup$
– Tigran Minasyan
Mar 21 at 20:12
$begingroup$
@TigranMinasyan You basically did it already; see my edit.
$endgroup$
– J.G.
Mar 21 at 20:13
1
$begingroup$
Indeed. The CDF is $F(x)~=mathbf 1_0leq xlt 1~int_0^x c~(s^2+1)~mathrm d s+mathbf 1_1leq x\= tfrac 34(tfrac 13x^3+x)~mathbf 1_0leq xlt 1+mathbf 1_1leq x$
$endgroup$
– Graham Kemp
Mar 21 at 21:47
add a comment |
$begingroup$
The CDF cannot exceed $1$, but the PDF can. Your CDF is $frac34x+frac14x^3$ on $[0,,1]$, just as you calculated.
answered Mar 21 at 20:11
J.G.J.G.
32.6k23250
$endgroup$
The CDF cannot exceed $1$, but the PDF can. Your CDF is $frac34x+frac14x^3$ on $[0,,1]$, just as you calculated.
answered Mar 21 at 20:11
J.G.J.G.
32.6k23250
edited Mar 21 at 20:12
answered Mar 21 at 20:11
J.G.J.G.
32.6k23250
answered Mar 21 at 20:11
J.G.J.G.
32.6k23250
answered Mar 21 at 20:11
J.G.J.G.
32.6k23250
32.6k23250
$begingroup$
Then how can you find the CDF from the PDF?
$endgroup$
– Tigran Minasyan
Mar 21 at 20:12
$begingroup$
@TigranMinasyan You basically did it already; see my edit.
$endgroup$
– J.G.
Mar 21 at 20:13
1
$begingroup$
Indeed. The CDF is $F(x)~=mathbf 1_0leq xlt 1~int_0^x c~(s^2+1)~mathrm d s+mathbf 1_1leq x\= tfrac 34(tfrac 13x^3+x)~mathbf 1_0leq xlt 1+mathbf 1_1leq x$
$endgroup$
– Graham Kemp
Mar 21 at 21:47
add a comment |
$begingroup$
Then how can you find the CDF from the PDF?
$endgroup$
– Tigran Minasyan
Mar 21 at 20:12
$begingroup$
@TigranMinasyan You basically did it already; see my edit.
$endgroup$
– J.G.
Mar 21 at 20:13
1
$begingroup$
Indeed. The CDF is $F(x)~=mathbf 1_0leq xlt 1~int_0^x c~(s^2+1)~mathrm d s+mathbf 1_1leq x\= tfrac 34(tfrac 13x^3+x)~mathbf 1_0leq xlt 1+mathbf 1_1leq x$
$endgroup$
– Graham Kemp
Mar 21 at 21:47
$begingroup$
Then how can you find the CDF from the PDF?
$endgroup$
– Tigran Minasyan
Mar 21 at 20:12
$begingroup$
Then how can you find the CDF from the PDF?
$endgroup$
– Tigran Minasyan
Mar 21 at 20:12
$begingroup$
@TigranMinasyan You basically did it already; see my edit.
$endgroup$
– J.G.
Mar 21 at 20:13
$begingroup$
@TigranMinasyan You basically did it already; see my edit.
$endgroup$
– J.G.
Mar 21 at 20:13
1
1
$begingroup$
Indeed. The CDF is $F(x)~=mathbf 1_0leq xlt 1~int_0^x c~(s^2+1)~mathrm d s+mathbf 1_1leq x\= tfrac 34(tfrac 13x^3+x)~mathbf 1_0leq xlt 1+mathbf 1_1leq x$
$endgroup$
– Graham Kemp
Mar 21 at 21:47
$begingroup$
Indeed. The CDF is $F(x)~=mathbf 1_0leq xlt 1~int_0^x c~(s^2+1)~mathrm d s+mathbf 1_1leq x\= tfrac 34(tfrac 13x^3+x)~mathbf 1_0leq xlt 1+mathbf 1_1leq x$
$endgroup$
– Graham Kemp
Mar 21 at 21:47
add a comment |
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