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 ~


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 $$f_y(y)=\frac 12 e^{-y/2}$$ $$P(Y>2)=\frac 12\int_2^\infty e^{-y/2} dy=-\left |e^{-y/2} \right |_2^\infty=\frac 1e$$

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.