Find the cumulative distribution function given a pdf

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A distribution X is described by the probability density function



$f(x) = fracxalpha^2e^-fracx^22alpha^2$, where $xge0.$



Find the cumulative distribution function.



What I have so far is $F(x) = 1-e^-fracx^22alpha^2$ for $xge0$ using the integral
$int^x_0fractalpha^2e^-fract^22alpha^2dt$ .



Is this all I need? I feel like I'm missing something...




probability statistics probability-distributions




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  • But $xge0$ though so why should that be considered? Confused
    – Sonjov
    Aug 21 at 14:12










  • Oops. You're right.
    – Eric Towers
    Aug 21 at 14:13






  • 1




    They asked you to find the CDF, and you found it! You are done.
    – Mike Earnest
    Aug 21 at 14:54














up vote
1
down vote

favorite












A distribution X is described by the probability density function



$f(x) = fracxalpha^2e^-fracx^22alpha^2$, where $xge0.$



Find the cumulative distribution function.



What I have so far is $F(x) = 1-e^-fracx^22alpha^2$ for $xge0$ using the integral
$int^x_0fractalpha^2e^-fract^22alpha^2dt$ .



Is this all I need? I feel like I'm missing something...




probability statistics probability-distributions




share|cite|improve this question




















  • But $xge0$ though so why should that be considered? Confused
    – Sonjov
    Aug 21 at 14:12










  • Oops. You're right.
    – Eric Towers
    Aug 21 at 14:13






  • 1




    They asked you to find the CDF, and you found it! You are done.
    – Mike Earnest
    Aug 21 at 14:54












up vote
1
down vote

favorite









up vote
1
down vote

favorite











A distribution X is described by the probability density function



$f(x) = fracxalpha^2e^-fracx^22alpha^2$, where $xge0.$



Find the cumulative distribution function.



What I have so far is $F(x) = 1-e^-fracx^22alpha^2$ for $xge0$ using the integral
$int^x_0fractalpha^2e^-fract^22alpha^2dt$ .



Is this all I need? I feel like I'm missing something...




probability statistics probability-distributions




share|cite|improve this question












A distribution X is described by the probability density function



$f(x) = fracxalpha^2e^-fracx^22alpha^2$, where $xge0.$



Find the cumulative distribution function.



What I have so far is $F(x) = 1-e^-fracx^22alpha^2$ for $xge0$ using the integral
$int^x_0fractalpha^2e^-fract^22alpha^2dt$ .



Is this all I need? I feel like I'm missing something...





probability statistics probability-distributions





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 21 at 14:07









Sonjov

1086




1086











  • But $xge0$ though so why should that be considered? Confused
    – Sonjov
    Aug 21 at 14:12










  • Oops. You're right.
    – Eric Towers
    Aug 21 at 14:13






  • 1




    They asked you to find the CDF, and you found it! You are done.
    – Mike Earnest
    Aug 21 at 14:54
















  • But $xge0$ though so why should that be considered? Confused
    – Sonjov
    Aug 21 at 14:12










  • Oops. You're right.
    – Eric Towers
    Aug 21 at 14:13






  • 1




    They asked you to find the CDF, and you found it! You are done.
    – Mike Earnest
    Aug 21 at 14:54















But $xge0$ though so why should that be considered? Confused
– Sonjov
Aug 21 at 14:12




But $xge0$ though so why should that be considered? Confused
– Sonjov
Aug 21 at 14:12












Oops. You're right.
– Eric Towers
Aug 21 at 14:13




Oops. You're right.
– Eric Towers
Aug 21 at 14:13




1




1




They asked you to find the CDF, and you found it! You are done.
– Mike Earnest
Aug 21 at 14:54




They asked you to find the CDF, and you found it! You are done.
– Mike Earnest
Aug 21 at 14:54










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










Assuming $alpha neq 0$, your answer is correct. (If $alpha = 0$, the PDF is undefined, so there's not much to do there either.)






share|cite|improve this answer




















  • Does it change anything if the distribution is only defined when $alpha^2>0$?
    – Sonjov
    Aug 21 at 14:17










  • @Sonjov : Notice that for $alpha in mathbbR$, $alpha^2 > 0$ and $alpha neq 0$ are equivalent.
    – Eric Towers
    Aug 21 at 14:18











  • Ah of course. Thanks.
    – Sonjov
    Aug 21 at 14:19










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Assuming $alpha neq 0$, your answer is correct. (If $alpha = 0$, the PDF is undefined, so there's not much to do there either.)






share|cite|improve this answer




















  • Does it change anything if the distribution is only defined when $alpha^2>0$?
    – Sonjov
    Aug 21 at 14:17










  • @Sonjov : Notice that for $alpha in mathbbR$, $alpha^2 > 0$ and $alpha neq 0$ are equivalent.
    – Eric Towers
    Aug 21 at 14:18











  • Ah of course. Thanks.
    – Sonjov
    Aug 21 at 14:19














up vote
2
down vote



accepted










Assuming $alpha neq 0$, your answer is correct. (If $alpha = 0$, the PDF is undefined, so there's not much to do there either.)






share|cite|improve this answer




















  • Does it change anything if the distribution is only defined when $alpha^2>0$?
    – Sonjov
    Aug 21 at 14:17










  • @Sonjov : Notice that for $alpha in mathbbR$, $alpha^2 > 0$ and $alpha neq 0$ are equivalent.
    – Eric Towers
    Aug 21 at 14:18











  • Ah of course. Thanks.
    – Sonjov
    Aug 21 at 14:19












up vote
2
down vote



accepted







up vote
2
down vote



accepted






Assuming $alpha neq 0$, your answer is correct. (If $alpha = 0$, the PDF is undefined, so there's not much to do there either.)






share|cite|improve this answer












Assuming $alpha neq 0$, your answer is correct. (If $alpha = 0$, the PDF is undefined, so there's not much to do there either.)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 21 at 14:15









Eric Towers

30.6k22264




30.6k22264











  • Does it change anything if the distribution is only defined when $alpha^2>0$?
    – Sonjov
    Aug 21 at 14:17










  • @Sonjov : Notice that for $alpha in mathbbR$, $alpha^2 > 0$ and $alpha neq 0$ are equivalent.
    – Eric Towers
    Aug 21 at 14:18











  • Ah of course. Thanks.
    – Sonjov
    Aug 21 at 14:19
















  • Does it change anything if the distribution is only defined when $alpha^2>0$?
    – Sonjov
    Aug 21 at 14:17










  • @Sonjov : Notice that for $alpha in mathbbR$, $alpha^2 > 0$ and $alpha neq 0$ are equivalent.
    – Eric Towers
    Aug 21 at 14:18











  • Ah of course. Thanks.
    – Sonjov
    Aug 21 at 14:19















Does it change anything if the distribution is only defined when $alpha^2>0$?
– Sonjov
Aug 21 at 14:17




Does it change anything if the distribution is only defined when $alpha^2>0$?
– Sonjov
Aug 21 at 14:17












@Sonjov : Notice that for $alpha in mathbbR$, $alpha^2 > 0$ and $alpha neq 0$ are equivalent.
– Eric Towers
Aug 21 at 14:18





@Sonjov : Notice that for $alpha in mathbbR$, $alpha^2 > 0$ and $alpha neq 0$ are equivalent.
– Eric Towers
Aug 21 at 14:18













Ah of course. Thanks.
– Sonjov
Aug 21 at 14:19




Ah of course. Thanks.
– Sonjov
Aug 21 at 14:19












 

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