Major League Baseball’s 2019 Most Valuable Player Awards officially were announced Thursday, ending the speculation surrounding the coveted award.

Los Angeles Angels center fielder Mike Trout won the award in the American League while Los Angeles Dodgers outfielder Cody Bellinger received the National League honor.

Trout beat out Houston Astros third baseman Alex Bregman and Oakland Athletics shortstop Marcus Semien to capture his third MVP award. He received 17 first-place votes and 13 second-place votes. Bregman finished in second with 13 first-place votes and 17 second-place votes, while Semien received zero first or second-place votes.

Bellinger beat out 2018 NL MVP and Milwaukee Brewers right fielder Christian Yelich and Washington Nationals third baseman Anthony Rendon. Bellinger received 19 first-place votes, 10 second-place votes, and a surprising single fifth-place vote, while Yelich finished with 10 first-place votes, 18 second-place votes, one third and one fourth. Rendon received one first and second-place vote.

Trout, 28, earns the honor after finishing in second in 2018 and fourth in 2017 since his last MVP award in 2016. He racked up a .291 batting average, 45 home runs, 104 RBIs, and an AL-leading 1.083 OPS in just 134 games in 2019, according to Fangraphs. Trout now has the second most MVPs in history just behind Barry Bonds with nine.

Bellinger, 24, earns the honor in just his second full MLB season, third overall and just two seasons after winning the NL Rookie of the Year. Bellinger’s stats were out of this world, as he slugged his way to .305 batting average, 47 home runs, 115 RBIs and a 1.035 OPS, ranking in the top 10 of each category in the NL.

For those wondering, Boston Red Sox shortstop Xander Bogaerts finished fifth in the voting and Mookie Betts finished in eighth.

Thumbnail photo via Erik Williams/USA TODAY Sports