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Let $X$ be normal with mean $0$ and variance $1/2$, Y exponential with mean $1$. $X$ and $Y$ are independent. Find $\mathbb{P}(Y>X^2)$

  • $\mathbb{P}(Y>X^2)=\int\limits_{-\infty}^{\infty}\mathbb{P}(Y>X^2|X=x)f_{X}(x)dx= ...=\sqrt{2}/2$

  • Second approach. Since $2X^2$ is $\Gamma(1/2,1/2)$, then $2\cdot2X^2$ is $\Gamma(1,1/2)$ which is exponential with mean $2$. Thus $\mathbb{P}(Y>X^2)=P(4Y>4X^2)=P(Y>0.25T)=a\neq\frac{\sqrt{2}}{2}$, where $T$ is exponentially distributed, indepedent of $Y$.

The question is: where am I wrong? I suspect the second approach is wrong, but why?

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$2*2X^2\sim\Gamma(1/2,1)$ if $2X^2\sim\Gamma(1/2,1/2)$ –  Seyhmus Güngören Jun 8 '13 at 20:21

1 Answer 1

More generally: given $X$~$N(\mu,\sigma^2)$ and $Y$~Exponential(1), with $X$ and $Y$ independent, their joint pdf is $f(x,y)$:

Here is a quick check of the probability you seek, using the mathStatica add-on to Mathematica:

Then, for your given parameter values, the correct solution is:

which is approximately 0.7.


Finally, a quick Monte Carlo check to confirm we haven't made any mistakes ... (the following sets $\mu=0$ and $\sigma = \sqrt\frac12$):

datax = RandomReal[NormalDistribution[0, Sqrt[1/2]], 100000]; 
datay = RandomReal[ExponentialDistribution[1], 100000]; 
Count[datay - datax^2, x_ /; x > 0] /100000.

0.70418

:)


OP wrote: - Second approach. "Since $2X^2$ is $\Gamma(1/2,1/2)$ ..."

Correctly: Since $\sqrt{2}X$ ~ $N(0,1)$, and the square of a standard Normal is Chisquared(1), it follows that: $2X^2$ ~ Chisquared(1) ... which is: Gamma$(\frac12,2)$ ... not Gamma$(\frac12,\frac12)$.

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Quote: The question is: where am I wrong? –  Did Jun 9 '13 at 16:18
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@Did The answer is: everywhere –  wolfies Jun 9 '13 at 16:50
    
Then, how is your post describing/identifying/remedying to, this "everywhere"? That is, how is it answering the question? –  Did Jun 9 '13 at 17:37
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@Did The OP's question title is: Find the probability. This I have done. If you would like to address his second-level question as to how/where he went wrong, the site does allow (and in fact encourages) others to submit solutions that may be helpful, and I would certainly encourage you to do so. Though, as this may possibly be a homework assignment, you might want to take care not to unduly interfere with the parent university's ability to carry out proper assessments. –  wolfies Jun 9 '13 at 17:57
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As @Did probably knows, displaying a CAS calculation and Monte Carlo simulation that agree with the OP's first calculation and disagree with the second, answers part of "where am I wrong" and confirms the "I suspect the second approach is wrong". It is wasting everyone's time to post a potentially endless sequence of negative comments in which one user tries to control the interpretation of words like question and answer in a quixotic campaign to demonstrate that this (useful) answer and ones like it are not a valid response to the OP's question, or are "not mathematical". –  zyx Jun 10 '13 at 3:13

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