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Trying to understand the following integral used to find the marginal probability density function for continuous random variables x & y. I haven't taken a calculus class in 20+ years, so this is a bit beyond me. I get confused w/ exponentials involving $e$, and integration by parts I'm maybe not doing correctly, but I don't know where the problem is.

$$f_y(y)=\int_0^\infty1/8xe^{-(x+y)/2}dx $$

Since I'm integrating w/respect to x, I think I should be able to do something like:

$$f_y(y)=(1/8)e^{-y/2}\int_0^\infty xe^{-x/2}dx $$

Is this correct for splitting up the exponential part or is this wrong? Seemed easier than messing with u-substitution, which got confusing w/int by parts as well...

So this w/integration by parts:

$u=x, du=dx, dv=e^{-x/2}, v=\frac {-1}2e^{-x/2}$

(is this how to integrate $e$? Or do I have it backwards? Should it be $v=-2e^{-x/2}$ instead? Tried it both ways w/2nd approach below. Neither gave me the right answer...

$$=(1/8)e^{-y/2} ( -2xe^{-x/2} |_0^\infty - \int_0^\infty 2e^{-x/2}dx )$$

$$=(1/8)e^{-y/2} ( -2xe^{-x/2}|_0^\infty - (2e^{-\infty} + 2e^0))$$

$$=(1/8)e^{-y/2} ( -2 ) $$ $$=\frac {-1}{4}e^{-y/2} $$

I have no idea if this is right. Doesn't seem likely.

I do know the answer to $P(Y > 2)$ is supposed to be $e^{-1}$

This should be $f_y(y)$ from 2 to infinity. Except I can't get that right either, again because I'm fuzzy on how $e$ integration works ...

$$=\int_2^\infty \frac {-1}{4}e^{-y/2} dy$$

$$=\frac {1}{2}e^{-y/2} |_2^\infty$$

$$=(\frac {1}{2}e^{-\infty}) - (\frac {1}{2}e^{-1})$$

$$=\frac {-1}{2}e^{-1}$$

Wrong ... I always wind up with some extra fraction out front (1/2, 1/4, or 1/8, or 1/16 depending what I try), so clearly there's some integration step I'm missing or have misunderstood.

(Also, highly likely there are typos or basic algebra mistakes above. First time typing anything on here & I was editing as I went ...)

Since I can't get the marginal PDF right to begin with, knowing the right answer is kind of moot.

A later question requires finding $f_x(x)$, and I know that $f_x(x) * f_y(y) = f(x,y)$ in this case, but I have the same difficulty w/the mechanics of it - I just can't do the integration.

Any chance somebody competent could help walk through this to highlight specifically where I'm going wrong (step-by-step, for dummies please?) Thanks ~

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you made a sign mistake when you performed integration by part $$I=(1/8)e^{-y/2} ( -2xe^{-x/2} |_0^\infty - \int_0^\infty 2e^{-x/2}dx )$$ It should be $$I=(1/8)e^{-y/2} ( -2xe^{-x/2} |_0^\infty \color{red}{+} \int_0^\infty 2e^{-x/2}dx )$$

And you have a coefficient mistake here too $$I=\int_0^\infty 2e^{-x/2}dx )=-4 \left | e^{-x/2} \right |_0^\infty$$

Because $$I=\int e^{ax}dx=\frac 1ae^{ax}$$

Once you correct your integration mistakes you get $$f_y(y)=\frac 12 e^{-y/2}$$ $$P(Y>2)=\frac 12\int_2^\infty e^{-y/2} dy=-\left |e^{-y/2} \right |_2^\infty=\frac 1e$$

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    $\begingroup$ Perfect, thanks! Knew it would be something simple like that, but I can never keep track of what to do w/$e$ ... And I have apparently been taking its derivative when I meant to be integrating. $\endgroup$ – mc01 Jun 26 '18 at 16:08
  • $\begingroup$ yes you did a good job just some little mistakes for the integral of $e^{ax}$ like you said you confused with the derivative. $\endgroup$ – Isham Jun 26 '18 at 16:10

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